A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is ______ J.
Explanation:
In a P-V diagram (Pressure vs. Volume), the total work done during a cyclic process is equivalent to the area enclosed by the cycle.
The cycle $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{A}$ is moving in a clockwise direction.
In a clockwise cycle on a P-V diagram, the work done by the system is positive.
The area enclosed by the triangle ABC is calculated using the formula for the area of a right-angled triangle:
$ \begin{aligned} & \text { Work Done }=\text { Area }=\frac{1}{2} \times \text { Base × Height } \\ & \text { Base }=V_C-V_A=5 \mathrm{~m}^3-2 \mathrm{~m}^3=3 \mathrm{~m}^3 \\ & \text { Height }=P_B-P_C=300 \mathrm{~Pa}-100 \mathrm{~Pa}=200 \mathrm{~Pa} \end{aligned} $
Substituting the values into the area formula:
$ \text { Work Done }=\frac{1}{2} \times 3 \mathrm{~m}^3 \times 200 \mathrm{~Pa}=300 \mathrm{~J} $
Therefore, the total work done by the system during the cycle ABC is 300 J . Hence, the correct answer is $\mathbf{3 0 0}$.
When 300 J of heat given to an ideal gas with $C_p=\frac{7}{2} R$ its temperature raises from $20^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ keeping its volume constant. The mass of the gas is (approximately) $\_\_\_\_$ g. $(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K})$
Explanation:
For a constant volume process, the heat supplied ( Q ) is given by the formula :
$ \mathrm{Q}=\mathrm{n} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{~T} $
Where :
1. $\mathrm{n}=$ number of moles of the gas.
2. $ \mathrm{C}_{\mathrm{v}}=$ molar specific heat capacity at constant volume.
3. $\Delta \mathrm{T}=$ change in temperature $50^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=30^{\circ} \mathrm{C}=30 \mathrm{~K}$
We are given the molar specific heat at constant pressure, $C_p=\frac{7}{2} R$. Using Mayer's Relation $\left(C_p-C_v=R\right)$ :
$ C_v=C_p-R $
$\Rightarrow $ $ C_v=\frac{7}{2} R-R=\frac{5}{2} R $
Using the formula of heat,
$ 300=\mathrm{n} \times\left(\frac{5}{2} \times 8.314\right) \times 30 $
$ \mathrm{n}=\frac{300 \times 2}{5 \times 8.314 \times 30} \approx 0.481 \mathrm{moles} $
So, the amount of the gas is 0.481 moles. As the molar mass of the gas is not given in the question, we can't determine the mass of the gas directly.
A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature $50^{\circ} \mathrm{C}$. The gas is now heated to double its temperature. The modified pressure is $\_\_\_\_$ Pa .
Explanation:
The problem states the gas is filled in a closed cylinder. That means the volume of the gas remains constant during the heating process ( $\mathrm{V}=$ constant).
For a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute temperature.
$ \frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2} $
Initial pressure is $\mathrm{P}_1=3.23 \mathrm{kPa}=3.23 \times 1000 \mathrm{~Pa}=3230 \mathrm{~Pa}$
Initial temperature is $\mathrm{T}_1=50^{\circ} \mathrm{C}=(50+273) \mathrm{K}=323 \mathrm{~K}$
The final temperature is $\mathrm{T}_2=2 \times \mathrm{T}_1=2 \times 323 \mathrm{~K}=646 \mathrm{~K}$
Using the relation $\frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2}$ :
$ \mathrm{P}_2=\mathrm{P}_1 \times \frac{\mathrm{T}_2}{\mathrm{~T}_1} $
$\Rightarrow $ $P_2=3230 \times \frac{646}{323}$
$ \mathrm{P}_2=3230 \times 2=6460 \mathrm{~Pa} $
Therefore, the modified pressure is 6460 Pa . Hence, the correct answer is 6460 .
Other Solution :
$\begin{aligned} & \mathrm{V}=\text { constant } \\ & \text { so } \mathrm{P} \propto \mathrm{T} \\ & \mathrm{T}_{\mathrm{i}}=50^{\circ} \mathrm{C}=323 \mathrm{~K} \\ & \mathrm{~T}_{\mathrm{f}}=100^{\circ} \mathrm{C}=373 \mathrm{~K} \\ & \Rightarrow \frac{\mathrm{P}_{\mathrm{f}}}{\mathrm{P}_{\mathrm{i}}}=\frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}} \\ & \Rightarrow \frac{\mathrm{P}_{\mathrm{f}}}{3.23 \mathrm{kPa}}=\frac{373}{323} \\ & \Rightarrow \mathrm{P}_{\mathrm{f}}=3730 \mathrm{~Pa}\end{aligned}$
An insulated cylinder of volume $60 \mathrm{~cm}^3$ is filled with a gas at $27^{\circ} \mathrm{C}$ and 2 atmospheric pressure. Then the gas is compressed making the final volume as $20 \mathrm{~cm}^3$ while allowing the temperature to rise to $77^{\circ} \mathrm{C}$. The final pressure is $\_\_\_\_$ atmospheric pressure.
Explanation:
Initial state of the gas,
Initial Volume, $V_1=60 \mathrm{~cm}^3$
Initial Temperature, $\mathrm{T}_1=27^{\circ} \mathrm{C}=(27+273) \mathrm{K}=300 \mathrm{~K}$
Initial Pressure, $\mathrm{P}_1=2 \mathrm{~atm}$
Final state of the gas is,
Final Volume, $V_2=20 \mathrm{~cm}^3$
Final Temperature, $\mathrm{T}_2=77^{\circ} \mathrm{C}=(77+273) \mathrm{K}=350 \mathrm{~K}$
We need to calculate the pressure of the gas at final state.
Using Ideal gas equation, $\mathrm{PV}=\mathrm{nRT} \Rightarrow \frac{\mathrm{PV}}{\mathrm{T}}=\mathrm{nR}=$ constant
$ \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2} $
$\Rightarrow \mathrm{P}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1 \mathrm{~T}_2}{\mathrm{~T}_1 \mathrm{~V}_2}$
$\Rightarrow $ $P_2=\frac{2 \times 60 \times 350}{300 \times 20} \mathrm{~atm}$
$\Rightarrow $ $\mathrm{P}_2=7 \mathrm{~atm}$
Explanation:
An isobaric process occurs at constant pressure (P). For isobaric process :
The work done is $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}$
The change in internal energy is $\Delta \mathrm{U}=\mathrm{n} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}$
Heat added to the system is, $\mathrm{Q}=\mathrm{n} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}$
The ratio of specific heats is defined as $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{V}}}$.
Also, from Mayer's relation $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$.
The ratio of Heat to Work is:
$ \frac{Q}{W}=\frac{n C_p \Delta T}{n R \Delta T}=\frac{C_p}{R} $
Substituting $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ :
$ \frac{Q}{W}=\frac{C_p}{C_p-C_v} $
$\Rightarrow $ $ \frac{\mathrm{Q}}{\mathrm{~W}}=\frac{\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}}{\left(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\right)-1}=\frac{\gamma}{\gamma-1} $
The work done is given as, $\mathrm{W}=100 \mathrm{~J}$ and the ratio of specific heats $\gamma=1.4$ (for a diatomic gas) So,
$ \frac{\mathrm{Q}}{100}=\frac{1.4}{1.4-1} $
$\Rightarrow $ $\frac{Q}{100}=\frac{1.4}{0.4}$
$\Rightarrow $ $\frac{Q}{100}=3.5$
$\Rightarrow $ $ Q=350 \mathrm{~J} $
Therefore, the heat given to the gas is 350 J .
Hence, the correct option is $\mathbf{3 5 0}$.
10 mole of oxygen is heated at constant volume from $30^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The change in the internal energy of the gas is $\_\_\_\_$ cal. (The molecular specific heat of oxygen at constant pressure, $C_P=7 \mathrm{cal} / \mathrm{mol} .{ }^{\circ} \mathrm{C}$ and $\left.\mathrm{R}=2 \mathrm{cal} . / \mathrm{mol} .{ }^{\circ} \mathrm{C}.\right)$
Explanation:
For an ideal gas, according to Mayer's relation
$ C_p-C_v=R $
Where R is the universal gas constant.
$ C_v=C_p-R $
Given:
$\mathrm{C}_{\mathrm{p}}=7 \frac{\mathrm{cal}}{\mathrm{mol}^{\circ} \mathrm{C}}$
$\mathrm{R}=2 \frac{\mathrm{cal}}{\mathrm{mol}^{\circ} \mathrm{C}}$
So, the molecular specific heat of oxygen at constant volume is,
$ \mathrm{C}_{\mathrm{v}}=7-2=5 \frac{\mathrm{cal}}{\mathrm{~mol}^{\circ} \mathrm{C}} $
Internal energy change is a state function that depends only on the temperature change. For n moles of an ideal gas, it is given by the formula:
$ \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T} $
The given values are,
Number of moles $\mathrm{n}=10 \mathrm{~mol}$
Initial temperature, $\mathrm{T}_1=30^{\circ} \mathrm{C}$
Final temperature $\mathrm{T}_2=40^{\circ} \mathrm{C}$
So, the change in temperature is, $\Delta \mathrm{T}=\mathrm{T}_2-\mathrm{T}_1=40-30=10^{\circ} \mathrm{C}$
So, the change in internal energy is,
$ \Delta U=10 \times 5 \times 10=500 \mathrm{cal} $
Therefore, when 10 moles of oxygen is heated at constant volume from $30^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ internal energy increase by 500 cal.
Hence, the correct answer is 500 .
Explanation:
$\begin{aligned} & \mathrm{R}_1=\frac{\ell_1}{\mathrm{~K}_1 \mathrm{~A}_1}, \mathrm{R}_2=\frac{\ell_2}{\mathrm{~K}_2 \mathrm{~A}_2} \\ & \frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\Delta \mathrm{T}}{\mathrm{R}} \\ & \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=\left(\frac{\mathrm{dQ}}{\mu}\right)_2 \\ & \frac{400-\mathrm{T}}{\mathrm{R}_1}=\frac{\mathrm{T}-200}{\mathrm{R}_2} \\ & \frac{400-\mathrm{T}}{\mathrm{~T}-200}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\ell_1}{\ell_2}\right)\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2 \times \frac{\mathrm{K}_2}{\mathrm{~K}_1} \\ & =\frac{1}{2} \times\left(\frac{1}{2}\right)^2 \times 2 \\ & =\left(\frac{1}{4}\right) \end{aligned}$
$\begin{aligned} & \frac{400-\mathrm{T}}{\mathrm{~T}-200}=\frac{1}{4} \\ & 1600-4 \mathrm{~T}=\mathrm{T}-200 \\ & 5 \mathrm{~T}=1800 \\ & \mathrm{~T}=360 \mathrm{~K} \end{aligned}$
A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is $1727^{\circ} \mathrm{C}$ and power radiated by the wire is 94.2 W . Its emissivity is $\frac{x}{8}$ where $x=$ _________.
(Given $\sigma=6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}, \pi=3.14$ and assume that the emissivity of wire material is same at all wavelength.)
Explanation:
To determine the emissivity ($\varepsilon$) of the wire used in the bulb, we start with the given parameters and the formula for power radiated by a black body:
Length of the wire ($L$): 10 cm = 0.1 m
Diameter of the wire ($d$): 0.5 mm = 0.0005 m
Temperature of the wire ($T$): $1727^{\circ} \mathrm{C} = 2000 \mathrm{K}$
Power radiated ($P$): 94.2 W
The power radiated by the wire can be expressed as:
$ P = \varepsilon \sigma A T^4 $
Where:
$\sigma$ is the Stefan-Boltzmann constant ($6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}$)
$A$ is the surface area of the wire, calculated using the formula for the surface area of a cylinder: $A = \pi d L$
Substituting the given values into the formula:
$ 94.2 = \varepsilon \times (6 \times 10^{-8}) \times (3.14 \times 0.0005 \times 0.1) \times (2000)^4 $
Simplifying the expression inside the parentheses:
$ 94.2 = \varepsilon \times (6 \times 10^{-8}) \times 3.14 \times 0.0005 \times 0.1 \times (2000)^4 $
On solving for $\varepsilon$:
$ \varepsilon = \frac{94.2}{[(6 \times 10^{-8}) \times 3.14 \times 0.0005 \times 0.1 \times (2000)^4]} $
Given that $\varepsilon$ is $\frac{x}{8}$, evaluating this expression provides the value of $x = 5$. Thus, the emissivity $\varepsilon$ is $\frac{5}{8}$.
An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is ________ $\times 10^{-1} \mathrm{~J}$. (Take $\pi=3.14$ )
Explanation:
$\begin{aligned} &\text { Area of circle, } W=\frac{\pi}{4} d_1 d_2\\ &\begin{aligned} & \mathrm{W}=\frac{\pi}{4}(500-300) \times 10^3(350-150) \times 10^{-6} \\ & \mathrm{~W}=31.4 \mathrm{Joule} \\ & \mathrm{~W}=314 \times 10^{-1} \mathrm{Joule} \end{aligned} \end{aligned}$
The internal energy of air in $4 \mathrm{~m} \times 4 \mathrm{~m} \times 3 \mathrm{~m}$ sized room at 1 atmospheric pressure will be___________________$\times 10^6 \mathrm{~J}$
(Consider air as diatomic molecule)
Explanation:
To determine the internal energy of the air in a room that measures $4 \, \mathrm{m} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m}$ at 1 atmospheric pressure, we can follow these steps using the properties of a diatomic gas:
The volume of the room, $ V $, is calculated as:
$ V = 4 \, \mathrm{m} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m} = 48 \, \mathrm{m}^3 $
For a diatomic gas like air, the internal energy $ U $ is given by the formula:
$ U = nC_v T = n \frac{5RT}{2} $
Where:
$ n $ is the number of moles,
$ C_v $ is the molar heat capacity at constant volume,
$ R $ is the ideal gas constant, and
$ T $ is the temperature in Kelvin.
Using the ideal gas law, $ PV = nRT $, we can substitute for $ nRT $:
$ nRT = PV $
Therefore, the internal energy $ U $ becomes:
$ U = \frac{5}{2} PV $
Substituting the given values for atmospheric pressure $ P = 10^5 \, \mathrm{Pa} $ (1 atm) and volume $ V = 48 \, \mathrm{m}^3 $:
$ U = \frac{5}{2} \times 10^5 \, \mathrm{Pa} \times 48 \, \mathrm{m}^3 = 12 \times 10^6 \, \mathrm{J} $
Thus, the internal energy of the air in the room is $ 12 \times 10^6 \, \mathrm{J} $.
$\gamma_{\mathrm{A}}$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas $B$ having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If $\frac{\gamma_A}{\gamma_B}=\left(1+\frac{1}{n}\right)$, then the value of $n$ is ________ .
Explanation:
To determine the value of $ n $, we start by considering the specific heat ratios for gases A and B.
For gas A, which is monoatomic with 3 translational degrees of freedom:
The degrees of freedom, $ f_A $, is 3.
Therefore, $ \gamma_A = \frac{f_A + 2}{f_A} = \frac{3 + 2}{3} = \frac{5}{3} $.
For gas B, which is polyatomic with 3 translational, 3 rotational degrees of freedom, and 1 vibrational mode:
The total degrees of freedom, $ f_B $, is $ 3 + 3 + 2 \times 1 = 8 $ (since one vibrational mode contributes 2 degrees of freedom: 1 kinetic + 1 potential).
Therefore, $ \gamma_B = \frac{f_B + 2}{f_B} = \frac{8 + 2}{8} = \frac{10}{8} = \frac{5}{4} $.
Now, the relation given is:
$ \frac{\gamma_A}{\gamma_B} = \left(1 + \frac{1}{n}\right) $
Substitute the values of $ \gamma_A $ and $ \gamma_B $:
$ \frac{\frac{5}{3}}{\frac{5}{4}} = \frac{5}{3} \times \frac{4}{5} = \frac{20}{15} = \frac{4}{3} $
Equating to the given form:
$ \frac{4}{3} = 1 + \frac{1}{n} $
Thus, we have:
$ \frac{4}{3} - 1 = \frac{1}{n} $
$ \frac{1}{3} = \frac{1}{n} $
Solving for $ n $:
$ n = 3 $
Hence, the value of $ n $ is 3.
A container of fixed volume contains a gas at 27°C. To double the pressure of the gas, the temperature of gas should be raised to _____ °C.
Explanation:
To solve this problem, we can use the combined gas law in terms of pressure ($P$) and temperature ($T$). For a gas with a fixed volume, the relationship is given as:
$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $
where:
$ P_1 $ is the initial pressure,
$ T_1 $ is the initial temperature in Kelvin,
$ P_2 $ is the final pressure,
$ T_2 $ is the final temperature in Kelvin.
Given that the initial temperature $ T_1 = 27°C $, we need to convert it to Kelvin:
$ T_1 = 27°C + 273.15 = 300.15 \, \text{K} $
The final pressure $ P_2 $ is twice the initial pressure ($ P_2 = 2P_1 $). Substituting these values into the gas law equation gives:
$ \frac{P_1}{300.15} = \frac{2P_1}{T_2} $
Solving for $ T_2 $:
Cancel $ P_1 $ from both sides:
$ \frac{1}{300.15} = \frac{2}{T_2} $
Rearrange to solve for $ T_2 $:
$ T_2 = 2 \times 300.15 = 600.3 \, \text{K} $
Convert the final temperature back to Celsius:
$ T_2 = 600.3 \, \text{K} - 273.15 = 327.15°C $
Therefore, to double the pressure of the gas, the temperature should be raised to approximately 327°C.
The temperature of 1 mole of an ideal monoatomic gas is increased by $50^{\circ} \mathrm{C}$ at constant pressure. The total heat added and change in internal energy are $E_1$ and $E_2$, respectively. If $\frac{E_1}{E_2}=\frac{x}{9}$ then the value of $x$ is _________.
Explanation:
$ Q = n\,C_P\,\Delta T \quad \text{and} \quad \Delta U = n\,C_V\,\Delta T $
For a monoatomic ideal gas, the molar specific heats are given by:
$ C_V = \frac{3}{2}R \quad \text{and} \quad C_P = C_V + R = \frac{5}{2}R $
Thus, the ratio of the heat added to the change in internal energy is:
$ \frac{Q}{\Delta U} = \frac{C_P}{C_V} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} $
Given that:
$ \frac{E_1}{E_2} = \frac{x}{9} $
We equate the two ratios:
$ \frac{5}{3} = \frac{x}{9} $
Multiplying both sides by 9:
$ x = \frac{5}{3} \times 9 = 15 $
Thus, the value of $x$ is 15.
An ideal gas initially at $0^{\circ} \mathrm{C}$ temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is $3 / 2$, the change in temperature due to the thermodynamic process is _________ K.
Explanation:
We’re given an ideal gas initially at $T_i = 0^\circ \mathrm{C} = 273\,\mathrm{K}.$ The gas is suddenly compressed to one-fourth of its initial volume, and we are told that
$\gamma = \frac{c_p}{c_v} = \frac{3}{2}\,.$
For an adiabatic process (one in which there is no heat exchange), if the process were reversible, the relation between the temperature and volume would be
$T\, V^{\gamma-1} = \text{constant}\,.$
Even though the compression is “sudden” (and hence irreversible), the final equilibrium state of the gas is uniquely determined by its internal energy (which is a state function). That allows us to use the adiabatic relation between the initial and final states.
Here are the steps:
Write the adiabatic relation between the initial and final states:
$T_i\, V_i^{\gamma-1} = T_f\, V_f^{\gamma-1}\,.$
Since the gas is compressed to one-fourth of its volume, we have
$V_f = \frac{1}{4} V_i\,.$
Substitute this into the adiabatic relation:
$273\,V_i^{\gamma-1} = T_f \left(\frac{1}{4} V_i\right)^{\gamma-1}\,.$
With $\gamma = \frac{3}{2},$
$\gamma - 1 = \frac{3}{2} - 1 = \frac{1}{2}\,.$
So the relation becomes:
$273\,V_i^{1/2} = T_f \left(\frac{1}{4}\right)^{1/2} V_i^{1/2}\,.$
Cancel the common factor $V_i^{1/2}$ (provided it is nonzero) and simplify:
$273 = T_f \left(\frac{1}{2}\right)\,.$
Now solve for the final temperature $T_f$:
$T_f = 273 \times 2 = 546\,\mathrm{K}\,.$
The change in temperature is then:
$\Delta T = T_f - T_i = 546 - 273 = 273\,\mathrm{K}\,.$
So, the change in temperature due to the thermodynamic process is $\boxed{273\,\mathrm{K}}.$
Three conductors of same length having thermal conductivity $k_1, k_2$ and $k_3$ are connected as shown in figure.
Area of cross sections of $1^{\text {st }}$ and $2^{\text {nd }}$ conductor are same and for $3^{\text {rd }}$ conductor it is double of the $1^{\text {st }}$ conductor. The temperatures are given in the figure. In steady state condition, the value of $\theta$ is _________ ${ }^{\circ} \mathrm{C}$. (Given : $\mathrm{k}_1=60 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}, \mathrm{k}_2=120 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}, \mathrm{k}_3=135 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ )
Explanation:
${R_1} = {{2L} \over {{K_1}A}}$
(As ${l_1} = {l_2} = {l_3} = l$
${A_1} = {A_2} = A/2$
${A_3} = A$)
${R_2} = {{2L} \over {{K_2}A}}$
${R_3} = {L \over {{K_3}A}}$
${I_1} + {I_2} = {I_3}$
$ \Rightarrow {{100 - \theta } \over {{{2L} \over {{K_1}A}}}} + {{100 - \theta } \over {{{2L} \over {{K_2}A}}}} = {{\theta - 0} \over {{L \over {{K_3}A}}}}$
$ \Rightarrow {{100 - \theta } \over 2}({K_1} + {K_2}) = {K_3}\theta $
$ \Rightarrow 50({K_1} + {K_2}) = \theta \left[ {{K_3} + {{{K_1} + {K_2}} \over 2}} \right]$
$ \Rightarrow 50(60 + 120) = \theta \left[ {135 + {{60 + 120} \over 2}} \right]$
$ \Rightarrow 50 \times 80 = \theta (135 + 90)$
$ \Rightarrow \theta = {{50 \times 180} \over {225}}$
$ \Rightarrow \theta = 40^\circ C$
Two plates $\mathrm{A}$ and $\mathrm{B}$ have thermal conductivities $84 ~\mathrm{Wm}^{-1} \mathrm{~K}^{-1}$ and $126 ~\mathrm{Wm}^{-1} \mathrm{~K}^{-1}$ respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of $\mathrm{A}$ and $\mathrm{B}$ are kept at $100^{\circ} \mathrm{C}$ and $0{ }^{\circ} \mathrm{C}$ respectively, then the temperature of the surface of contact in steady state is _____________ ${ }^{\circ} \mathrm{C}$.
Explanation:
We can use the formula for heat transfer rate through a plate:
$Q = kA\frac{T_2 - T_1}{d}$
Where Q is the heat transfer rate, k is the thermal conductivity, A is the surface area, T1 and T2 are the temperatures on either side of the plate, and d is the thickness of the plate.
For plate A:
$Q_A = k_A A\frac{T_A - T}{d}$
For plate B:
$Q_B = k_B A\frac{T - T_B}{d}$
Since the heat transfer rate is the same through both plates in steady state:
$Q_A = Q_B$
We can now substitute the given values for thermal conductivities and temperatures:
$84A\frac{100 - T}{d} = 126A\frac{T - 0}{d}$
Notice that the surface area (A) and thickness (d) are the same for both plates, so they cancel out:
$84(100 - T) = 126T$
Now, we can solve for T:
$8400 - 84T = 126T$
$210T = 8400$
$T = \frac{8400}{210}$
$T = 40$
So, the temperature of the surface of contact in steady state is $40{ }^{\circ} \mathrm{C}$.
A steel rod of length $1 \mathrm{~m}$ and cross sectional area $10^{-4} \mathrm{~m}^{2}$ is heated from $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$ without being allowed to extend or bend. The compressive tension produced in the rod is ___________ $\times 10^{4} \mathrm{~N}$. (Given Young's modulus of steel $=2 \times 10^{11} \mathrm{Nm}^{-2}$, coefficient of linear expansion $=10^{-5} \mathrm{~K}^{-1}$ )
Explanation:
The change in length of the rod when it is heated is given by the equation:
$\Delta L = L_0 \cdot \alpha \cdot \Delta T$
where
- $\Delta L$ is the change in length,
- $L_0$ is the original length,
- $\alpha$ is the coefficient of linear expansion, and
- $\Delta T$ is the change in temperature.
Substituting the given values:
$\Delta L = 1 \, \text{m} \cdot 10^{-5} \, \text{K}^{-1} \cdot 200 \, \text{K} = 0.002 \, \text{m}$
The rod is not allowed to extend or bend, so a stress is created in the rod. This stress can be calculated using Young's modulus (Y), which is the ratio of the stress (force per unit area, F/A) to the strain (change in length per unit length, $\Delta L / L_0$):
$Y = \frac{F/A}{\Delta L / L_0}$
Rearranging for F gives:
$F = Y \cdot A \cdot \frac{\Delta L}{L_0}$
Substituting the given values:
$F = 2 \times 10^{11} \, \text{N/m}^2 \cdot 10^{-4} \, \text{m}^2 \cdot \frac{0.002 \, \text{m}}{1 \, \text{m}} = 4 \times 10^{4} \, \text{N}$
So the compressive tension produced in the rod is $4 \times 10^{4} \, \text{N}$.
(Assume that the specific heat capacity of water remains constant over the temperature range of the water).
Explanation:
The amount of heat energy required to raise the temperature of a substance can be calculated as:
Q = m $ \times $ c $ \times $ ΔT
where Q is the heat energy required, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
The time required to heat a substance can be calculated as :
t = ${Q \over P}$
where t is the time required, and P is the power of the heating device.
The actual power output of the heating device can be calculated as:
Pactual = Pinput $ \times $ efficiency
where P_input is the input power to the device and efficiency is the fraction of input power that is actually converted to useful power output.
Substituting the given values:
Q = 2 kg $ \times $ 4200 J/kg/K $ \times $ (60 - 10) = 2 kg $ \times $ 4200 J/kg/K $ \times $ 50 K = 4200 $ \times $ 50 $ \times $ 2 J = 420,000 J
Pinput = 2000 W = 2000 J/s
Pactual = 2000 $ \times $ 0.7 = 1400 J/s
t = ${Q \over {{P_{actual}}}}$ = ${{420,000} \over {1400}}$ J/s = 300 s
So, the time required to heat 2 kg of water from 10°C to 60°C is approximately 300 s.
Explanation:
Let the correct temperature be X$^\circ$C
$ \Rightarrow {{X - 0} \over {100 - 0}} = {{41 - 5} \over {95 - 5}} \Rightarrow X = 40$
$\Rightarrow$ Temperature is 273 + 40 K = 313 K
A hole is drilled in a metal sheet. At $\mathrm{27^\circ C}$, the diameter of hole is 5 cm. When the sheet is heated to $\mathrm{177^\circ C}$, the change in the diameter of hole is $\mathrm{d\times10^{-3}}$ cm. The value of d will be __________ if coefficient of linear expansion of the metal is $1.6\times10^{-5}/^\circ$C.
Explanation:
$ \begin{aligned} & =5 \times 1.6 \times 10^{-5}(177-27) \\\\ & =0.012 \mathrm{~cm} \\\\ & =12 \times 10^{-3} \mathrm{~cm} \\\\ & \text { so, } d=12 \end{aligned} $
The pressure $\mathrm{P}_{1}$ and density $\mathrm{d}_{1}$ of diatomic gas $\left(\gamma=\frac{7}{5}\right)$ changes suddenly to $\mathrm{P}_{2}\left(>\mathrm{P}_{1}\right)$ and $\mathrm{d}_{2}$ respectively during an adiabatic process. The temperature of the gas increases and becomes ________ times of its initial temperature. (given $\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}=32$)
Explanation:
${P_1}V_1^\gamma = {P_2}V_2^2$
${{{P_1}} \over {d_1^\gamma }} = {{{P_2}} \over {d_2^\gamma }}$
${{{d_1}{T_1}} \over {d_1^\gamma }} = {{{d_2}{T_2}} \over {d_2^\gamma }}$
${T_2} = {\left( {{{{d_2}} \over {{d_1}}}} \right)^{\gamma - 1}}{T_1}$
$ = {(32)^{{2 \over 5}}}{T_1}$
${T_2} = 4\,{T_1}$
One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is $\frac{\alpha^{2}}{4} \mathrm{R} \,\mathrm{J} / \mathrm{mol} \,\mathrm{K}$; then the value of $\alpha$ will be _________. (Assume that the given diatomic gas has no vibrational mode).
Explanation:
${C_V} = {f \over 2}R$
total degree of freedoms
$ = 1 \times 3 + 3 \times 5 = 18$
${{{\alpha ^2}} \over 4} = {{18} \over {2n}} = {{18} \over {2 \times 4}}$
$ \Rightarrow {\alpha ^2} = 9$
$\alpha = 3$
At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J of work when it expands under constant pressure. The amount of heat absorbed by the gas will be _________ J.
Explanation:
$f = 8$
$W = P\,dV = 150$
$Q = W + \Delta U$
$ = P\,dV + {f \over 2}\,PdV$
$Q = 5 \times 150 = 750\,J$
A block of ice of mass 120 g at temperature 0$^\circ$C is put in 300 g of water at 25$^\circ$C. The x g of ice melts as the temperature of the water reaches 0$^\circ$C. The value of x is _____________.
[Use specific heat capacity of water = 4200 Jkg$-$1K$-$1, Latent heat of ice = 3.5 $\times$ 105 Jkg$-$1]
Explanation:
Heat lost by water = Heat gained by ice
$0.3 \times 4200 \times 25 = x \times 3.5 \times {10^5}$
$x = {{0.3 \times 4200 \times 25} \over {3.5 \times {{10}^5}}}$
$ = 90 \times 100 \times {10^5} \times {10^3}$ gram = 90 gm
A unit scale is to be prepared whose length does not change with temperature and remains $20 \mathrm{~cm}$, using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is $40 \mathrm{~cm}$ and length of iron will be __________ $\mathrm{cm}$. $\left(\alpha_{\text {iron }}=1.2 \times 10^{-5} \mathrm{~K}^{-1}\right.$ and $\left.\alpha_{\text {brass }}=1.8 \times 10^{-5} \mathrm{~K}^{-1}\right)$.
Explanation:
$\Delta {L_1} = {\alpha _1}{L_1}\Delta T$
$\Delta {L_2} = {\alpha _2}{L_2}\Delta T$
${\alpha _1}{L_1} = {\alpha _2}{L_2}$
$1.2 \times {10^{ - 5}} \times {L_1} = 1.8 \times {10^{ - 5}} \times {L_2}$
${L_1} = {{1.8} \over {1.2}} \times 40 = 60$ cm
Two coils require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be ___________ min.
Explanation:
$H = {{{V^2}} \over R}\,.\,\Delta t$
$ \Rightarrow H = {{{V^2}} \over {{R_1}}}\,.\,20 = {{{V^2}} \over {{R_2}}}\,.\,60$ ..... (i)
Also, $H = {{{V^2}} \over {\left[ {{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \right]}}\,.\,\Delta t$
$ = {4 \over 3}\,.\,{{{V^2}} \over {{R_1}}}\,.\,\Delta t$ [$\because$ ${R_2} = 3{R_1}$]
$ \Rightarrow \Delta t = 15$
As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm2 for each plate. The equivalent thermal conductivity of the compound plate is $\left( {1 + {5 \over \alpha }} \right)$ K, then the value of $\alpha$ will be ______________.
Explanation:
${{{L_1}} \over {{K_1}{A_1}}} + {{{L_2}} \over {{K_2}{A_2}}} = {{{L_1} + {L_2}} \over {{K_{eff}}{A_{eff}}}}$
$ \Rightarrow {4 \over K} + {{2.5} \over {2K}} = {{6.5} \over {{K_{eff}}}}$
$ \Rightarrow {{10.5} \over {2K}} = {{6.5} \over {{K_{eff}}}}$
$ \Rightarrow {K_{eff}} = {{13K} \over {10.5}} = \left( {1 + {5 \over {21}}} \right)K$
$ \Rightarrow \alpha = 21$
300 cal. of heat is given to a heat engine and it rejects 225 cal. If source temperature is 227$^\circ$C, then the temperature of sink will be ______________ $^\circ$C.
Explanation:
$\eta = {W \over Q} = {{300 - 225} \over {300}}$
$ \Rightarrow {{75} \over {300}} = 1 - {{{T_L}} \over {{T_H}}}$
$ \Rightarrow {T_L} = {3 \over 4}\,{T_H} = {3 \over 4}(500) = 375\,K$
$ \Rightarrow {T_L} = 102^\circ C$
The total internal energy of two mole monoatomic ideal gas at temperature T = 300 K will be _____________ J. (Given R = 8.31 J/mol.K)
Explanation:
$U = 2\left( {{3 \over 2}R} \right)300$
$ = 3 \times 8.31 \times 300$
$ = 7479$ J
A diatomic gas ($\gamma$ = 1.4) does 400J of work when it is expanded isobarically. The heat given to the gas in the process is __________ J.
Explanation:
W = nR$\Delta$T = 400 J
$\therefore$ $\Delta$Q = nCP$\Delta$T
$ = n \times {7 \over 2}R \times \Delta T = {7 \over 2} \times (400) = 1400$
In a carnot engine, the temperature of reservoir is 527$^\circ$C and that of sink is 200 K. If the work done by the engine when it transfers heat from reservoir to sink is 12000 kJ, the quantity of heat absorbed by the engine from reservoir is ______________ $\times$ 106 J.
Explanation:
$\eta = 1 - {{{T_2}} \over {{T_1}}}$
$ = 1 - {{200} \over {800}} = {3 \over 4}$
$\therefore$ $\eta = {W \over {{Q_1}}}$
$ \Rightarrow {3 \over 4} = {{12000 \times {{10}^3}} \over {{Q_1}}}$
$ \Rightarrow {Q_1} = 16 \times {10^6}\,J$
A geyser heats water flowing at a rate of 2.0 kg per minute from 30$^\circ$C to 70$^\circ$C. If geyser operates on a gas burner, the rate of combustion of fuel will be ___________ g min$-$1.
[Heat of combustion = 8 $\times$ 103 Jg$-$1, Specific heat of water = 4.2 Jg$-$1 $^\circ$C$-$1]
Explanation:
$Q = ms\Delta T$
${{dQ} \over {dt}} = {\left( {{{dm} \over {dt}}} \right)_{water}}S\Delta T = {\left( {{{dm} \over {dt}}} \right)_{oil}}C$
$ \Rightarrow 2 \times 4.2 \times {10^3} \times 40 = {\left( {{{dm} \over {dt}}} \right)_{oil}} \times 8 \times {10^6}$
$ \Rightarrow {\left( {{{dm} \over {dt}}} \right)_{oil}} = {{8 \times 4.2 \times {{10}^4}} \over {8 \times {{10}^6}}}$ kg/minute
= 42 g/min
A heat engine operates with the cold reservoir at temperature 324 K. The minimum temperature of the hot reservoir, if the heat engine takes 300 J heat from the hot reservoir and delivers 180 J heat to the cold reservoir per cycle, is ____________ K.
Explanation:
$\left( {1 - {{324} \over {{T_H}}}} \right) = {{300 - 180} \over {300}}$
$1 - {2 \over 5} = {{324} \over {{T_H}}}$
${T_H} = {{324 \times 5} \over 3} = 540$
When a gas filled in a closed vessel is heated by raising the temperature by 1$^\circ$C, its pressure increases by 0.4%. The initial temperature of the gas is ___________ K.
Explanation:
$PV = nRT$
So ${{dP} \over P} \times 100 = {{dT} \over T} \times 100$
$0.4 = {1 \over T} \times 100$
$ \Rightarrow T = 250\,K$
A steam engine intakes 50 g of steam at 100$^\circ$C per minute and cools it down to 20$^\circ$C. If latent heat of vaporization of steam is 540 cal g$-$1, then the heat rejected by the steam engine per minute is __________ $\times$ 103 cal.
(Given : specific heat capacity of water : 1 cal g$-$1 $^\circ$C$-$1)
Explanation:
$\Delta$Qrej = 50 $\times$ 540 + 50 $\times$ 1 $\times$ (100 $-$ 20)
= 50 $\times$ [540 + 80]
= 50 $\times$ 620
= 31000 cal
= 31 $\times$ 103 cal
A monoatomic gas performs a work of ${Q \over {4}}$ where Q is the heat supplied to it. The molar heat capacity of the gas will be ______________ R during this transformation. Where R is the gas constant.
Explanation:
By 1st law,
$\Delta U = \Delta Q - {{\Delta Q} \over 4} = {3 \over 4}\Delta Q$
$ \Rightarrow n{C_v}\Delta T = {3 \over 4}nC\Delta T$
$ \Rightarrow C = {{4{C_v}} \over 3} = 2R$
0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127$^\circ$C. Th amount of heat required to double the speed of its molecules is ____________ k cal.
Take R = 2 cal mole$-$1 K$-$1)
Explanation:
Because the vessel is closed, it will be an isochoric process.
To double the speed, temperature must be 4 times (v $\alpha$$\sqrt{T}$)
So, Tf = 1600 K, Ti = 400 K
number of moles are ${{56} \over {28}} = 2$
so Q = nCv $\Delta$T = 2 $\times$ ${5 \over 2}$ $\times$ 2 $\times$ 1200
= 12000 cal = 12 K cal
Explanation:
The increase in temperature of the diatomic mole,
$\Delta$T = 40$^\circ$C
Now, the degree of freedom
f = linear + rotational + no oscillation
f = 3 + 2 + 0 $\Rightarrow$ f = 5
Change in internal energy,
$\Delta$U = nCv$\Delta$T .... (i)
where, ${C_v} = {f \over 2}R = {5 \over 2}R$
Substituting the value in Eq. (i), we get
$\Delta U = {{5R} \over 2}nR\Delta T$
Now, work done by the gas for isobaric process,
$W = p\Delta V = nR\Delta T$
The ratio of the change in internal energy to the work done by the gas,
${{\Delta U} \over W} = {{{5 \over 2}nR\Delta T} \over {nR\Delta T}}$
$ = {{\Delta U} \over W} = {5 \over 2}$
Multiply and divide the above equation with 5, we get
${{\Delta U} \over W} = {{5 \times 5} \over {2 \times 5}} = {{25} \over {10}}$
Comparing with given equation, ${{\Delta U} \over W} = {x \over {10}}$
The value of the x = 25.
Explanation:
Translational kinetic energy of dinitrogen (N2)
$KE = {3 \over 2}{K_B}T$
Here, T = temperature of the gas,
and KB = Boltzmann constant.
Kinetic energy of an electron = eV
Given, the potential differential of an electron, V = 0.1 V
Substituting the values in the Eq. (i), we get
${3 \over 2}{K_B}T = eV$
$ \Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$
$T = 773K = 773 - 273^\circ C = 500^\circ C$
Explanation:
p1v1v = p2v2v
(200) (1200)1.5 = P2 (300)1.5
P2 = 200 [4]3/2 = 1600 kPa
$\left| {W.D.} \right| = {{{p_2}{v_2} - {p_1}{v_1}} \over {v - 1}} = \left( {{{480 - 240} \over {0.5}}} \right) = 480$ J
Explanation:
Work done = Qin $-$ Qout = 300 $-$ 240 = 60 J
Efficiency = ${W \over {{Q_{in}}}} = {{60} \over {300}} = {1 \over 5}$
efficiency = $1 - {{{T_2}} \over {{T_1}}}$
${1 \over 5} = 1 - {{400} \over {{T_1}}} \Rightarrow {{400} \over {{T_1}}} = {4 \over 5}$
T1 = 500 k
Explanation:
Rods are identical so
RAB = RCD = 10 Kw$-$1
C is mid-point of AB, so
RAC = RCB = 5 Kw$-$1
at point C
${{200 - T} \over 5} = {{T - 125} \over {10}} + {{T - 100} \over 5}$
2(200 $-$ T) = T $-$ 125 + 2(T $-$ 100)
400 $-$ 2T = T $-$ 125 + 2T $-$ 200
$T = {{725} \over 5}$ = 145$^\circ$C
${I_h} = {{145 - 125} \over {10}}w = {{20} \over {10}}w$
Ih = 2w
Explanation:
$\lambda = {1 \over {\sqrt 2 \pi {d^2}n}}$
${{{\lambda _1}} \over {{\lambda _2}}} = {{d_2^2{n_2}} \over {d_1^2{n_1}}}$
$ = {\left( {{5 \over {10}}} \right)^2} = 0.25 = 25 \times {10^{ - 2}}$
(Young's modulus of material of track is 1011 Nm$-$2)
Explanation:
${{(Stress)} \over y} = \alpha \Delta T$ or $\sigma = Y\alpha \Delta T$
Energy stored per unit volume = ${1 \over 2}{\sigma \over Y}$
$\Rightarrow$ Energy stored per unit length = ${{A{\sigma ^2}} \over {2Y}}$
$ = {A \over 2} \times Y{\alpha ^2}\Delta {T^2}$
$ = {{{{10}^{ - 2}} \times {{10}^{11}} \times {{10}^{ - 10}} \times 100} \over 2} = 5$ J/m
Explanation:
$W = nRT\ln \left( {{{{V_2}} \over {{V_1}}}} \right)$
= 1 $\times$ 8.3 $\times$ 300 $\times$ ln2
= 17258 $\times$ 10$-$1 J
Explanation:
Here, r1 = 10 $\times$ 103 m and r2 = 10 $\times$ 10-3 m
We know that for a complete cyclic process, change in internal energy, ($\Delta$U) = 0 .... (i)
According to 1st law of thermodynamics,
$\Delta$Q = $\Delta$U + W ..... (ii)
From Eqs. (i) and (ii), we get
$\Rightarrow$ $\Delta$Q = 0 + W
$\Rightarrow$ $\Delta$Q = W .... (iii)
$\because$ W = Area = $\pi$r1r2
= $\pi$ $\times$ (10 $\times$ 103) $\times$ (10 $\times$ 10$-$3)
W = 100 $\pi$ J .... (iv)
From Eqs. (iii) and (iv), we get
$\Delta$Q = 100 $\pi$ J
Explanation:
Given, temperature of source, TH = 127$^\circ$C
= 273 + 127 = 400 K
Efficiencies, $\eta $ = 60% = 0.6
The efficiency of Carnot (ideal) heat engine is given by
$\eta = \left( {1 - {{{T_L}} \over {{T_H}}}} \right)$
where, TL = temperature of sink,
$0.6 = \left( {1 - {{{T_L}} \over {{T_H}}}} \right) \Rightarrow {{{T_L}} \over {{T_H}}} = 1 - 0.6$
${{{T_L}} \over {{T_H}}} = 0.4 \Rightarrow {T_L} = 0.4 \times {T_H}$
$ = 0.4 \times 400 = 160K$
$ = 160 - 273 = - 113^\circ C$
Explanation:
$Q = \Delta U + W$
$Q = \Delta U + {Q \over 5}$
$\Delta U = {{4Q} \over 5}$
$n{C_v}\Delta T = {4 \over 5}nC\Delta T$
${5 \over 4}{C_v} = C$
$C = {5 \over 4}\left( {{f \over 2}} \right)R = {5 \over 4}\left( {{5 \over 2}} \right)R$
$C = {{25} \over 8}R$
$X = 25$
Explanation:
$W = \int {PdV} $ .... (1)
$ \Rightarrow P = {{nRT} \over V}$ .... (2)
$ \Rightarrow W = \int {{{nRT} \over V}dv} $ .... (3)
and given $V = K{T^{2/3}}$ .... (4)
$ \Rightarrow W = \int {{{nRT} \over {K{T^{2/3}}}}.dv} $ .... (5)
$ \Rightarrow $ from (4) : $dv = {2 \over 3}K{T^{ - 1/3}}dT$
$ \Rightarrow W = \int\limits_{{T_1}}^{{T_2}} {{{nRT} \over {K{T^{2/3}}}}{2 \over 3}K{1 \over {{T^{1/3}}}}} dT$
$ \Rightarrow W = {2 \over 3}nR \times \left( {{T_2} - {T_1}} \right)$ .... (6)
$ \Rightarrow {T_2} - {T_1} = 90K$ .... (7)
$ \Rightarrow W = {2 \over 3}nR \times 90$
$ \Rightarrow W = 60nR$
Assuming 1 mole of gas
n = 1
So, W = 60R
Explanation:
${3 \over 2}{n_1}R{T_1} + {3 \over 2}{n_2}R{T_2} = {3 \over 2}({n_1} + {n_2})RT$
Using PV = nRT
P1V1 + P2V2 = P(V1 + V2)
$P = {{{P_1}{V_1} + {P_2}{V_2}} \over {{V_1} + {V_2}}} = {{2 \times 4.5 + 3 \times 5.5} \over {4.5 + 5.5}}$
$P = {{9 + 16.5} \over {10}} = {{25.5} \over {10}}$
$ \approx 25 \times {10^{ - 1}}$ atm