Heat and Thermodynamics
The temperature of an ideal gas is increased from $200 \mathrm{~K}$ to $800 \mathrm{~K}$. If r.m.s. speed of gas at $200 \mathrm{~K}$ is $v_{0}$. Then, r.m.s. speed of the gas at $800 \mathrm{~K}$ will be:
A body cools in 7 minutes from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The temperature of the surrounding is $10^{\circ} \mathrm{C}$. The temperature of the body after the next 7 minutes will be:
The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is:
A source supplies heat to a system at the rate of $1000 \mathrm{~W}$. If the system performs work at a rate of $200 \mathrm{~W}$. The rate at which internal energy of the system increases is
The number of air molecules per cm$^3$ increased from $3\times10^{19}$ to $12\times10^{19}$. The ratio of collision frequency of air molecules before and after the increase in number respectively is:
A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$. When the temperature of cold reservoir raised by x, its efficiency decreases to $\frac{1}{6}$. The value of x, if the temperature of hot reservoir is $99^\circ$C, will be :
For three low density gases A, B, C pressure versus temperature graphs are plotted while keeping them at constant volume, as shown in the figure.
The temperature corresponding to the point '$\mathrm{K}$' is :
A sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is $\left(\mathrm{given}, \gamma=\frac{3}{2}\right)$ :
$\left(P+\frac{a}{V^{2}}\right)(V-b)=R T$ represents the equation of state of some gases. Where $P$ is the pressure, $V$ is the volume, $T$ is the temperature and $a, b, R$ are the constants. The physical quantity, which has dimensional formula as that of $\frac{b^{2}}{a}$, will be:
The average kinetic energy of a molecule of the gas is
The pressure of a gas changes linearly with volume from $\mathrm{A}$ to $\mathrm{B}$ as shown in figure. If no heat is supplied to or extracted from the gas then change in the internal energy of the gas will be
The correct relation between $\gamma = {{{c_p}} \over {{c_v}}}$ and temperature T is :
Assertion A: Efficiency of a reversible heat engine will be highest at $-273^{\circ} \mathrm{C}$ temperature of cold reservoir.
Reason R: The efficiency of Carnot's engine depends not only on the temperature of the cold reservoir but it depends on the temperature of the hot reservoir too and is given as $\eta=\left(1-\frac{T_{2}}{T_{1}}\right)$
In the light of the above statements, choose the correct answer from the options given below
The pressure $(\mathrm{P})$ and temperature ($\mathrm{T})$ relationship of an ideal gas obeys the equation $\mathrm{PT}^{2}=$ constant. The volume expansion coefficient of the gas will be :
Heat is given to an ideal gas in an isothermal process.
A. Internal energy of the gas will decrease.
B. Internal energy of the gas will increase.
C. Internal energy of the gas will not change.
D. The gas will do positive work.
E. The gas will do negative work.
Choose the correct answer from the options given below :
Heat energy of 184 kJ is given to ice of mass 600 g at $-12^\circ \mathrm{C}$. Specific heat of ice is $\mathrm{2222.3~J~kg^{-1^\circ}~C^{-1}}$ and latent heat of ice in 336 $\mathrm{kJ/kg^{-1}}$
A. Final temperature of system will be 0$^\circ$C.
B. Final temperature of the system will be greater than 0$^\circ$C.
C. The final system will have a mixture of ice and water in the ratio of 5 : 1.
D. The final system will have a mixture of ice and water in the ratio of 1 : 5.
E. The final system will have water only.
Choose the correct answer from the options given below :
At 300 K, the rms speed of oxygen molecules is $\sqrt {{{\alpha + 5} \over \alpha }} $ times to that of its average speed in the gas. Then, the value of $\alpha$ will be
(used $\pi = {{22} \over 7}$)
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: If $d Q$ and $d W$ represent the heat supplied to the system and the work done on the system respectively. Then according to the first law of thermodynamics $d Q=d U-d W$.
Reason R: First law of thermodynamics is based on law of conservation of energy.
In the light of the above statements, choose the correct answer from the options given below:
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Isothermal Process | I. | Work done by the gas decreases internal energy |
| B. | Adiabatic Process | II. | No change in internal energy |
| C. | Isochoric Process | III. | The heat absorbed goes partly to increase internal energy and partly to do work |
| D. | Isobaric Process | IV. | No work is done on or by the gas |
Choose the correct answer from the options given below :
The graph between two temperature scales P and Q is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale P and 100 divisions on scale Q. The relationship for conversion between the two scales is given by :-
According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-
The root mean square velocity of molecules of gas is
A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :
Let $\gamma_1$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $\gamma_2$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, $\frac{\gamma_1}{\gamma_2}$ is :
In an Isothermal change, the change in pressure and volume of a gas can be represented for three different temperature; $\mathrm{T_3 > T_2 > T_1}$ as :
1 g of a liquid is converted to vapour at 3 $\times$ 10$^5$ Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm$^3$ during this phase change, then the increase in internal energy in the process will be :
Given below are two statements :
Statement I : The temperature of a gas is $-73^\circ$C. When the gas is heated to $527^\circ$C, the root mean square speed of the molecules is doubled.
Statement II : The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.
In the light of the above statements, choose the correct answer from the option given below :
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} $
Explanation:
$\begin{aligned} W_1 & =W_a+W_b+W_c+W_d \\\\ & =4 P_0\left(2 V_0-V_0\right)+n R T \ln \left(\frac{4 V_0}{2 V_0}\right)+2 P_0\left(V_0-4 V_0\right)+0 \\\\ & =4 P_0 V_0+n R\left(\frac{8 P_0 V_0}{n R}\right) \ln 2-6 P_0 V_0 \\\\ & =8 P_0 V_0 \ln 2-2 P_0 V_0\end{aligned}$
$\begin{aligned} W_{\text {II }} & =W_a^{\prime}+W_b^{\prime}+W_c^{\prime}+W_d^{\prime} \\\\ & =n R T \ln \left(\frac{2 V_0}{V_0}\right)+0+P_0\left(V_0-2 V_0\right)+0 \\\\ & =n R\left(\frac{4 P_0 V_0}{n R}\right) \ln 2-P_0 V_0 \\\\ & =4 P_0 V_0 \ln 2-P_0 V_0\end{aligned}$
$\frac{W_I}{W_{I I}}=\frac{8 P_0 V_0 \ln 2-2 P_0 V_0}{4 P_0 V_0 \ln 2-P_0 V_0}=2$
Explanation:
For a gas R and M are constant. So, $\rho T$ = Constant (for constant pressure).
The density of hot air inside the furnace is = $\rho $
The air gets heated inside the furnace at constant pressure Pa.
$ \therefore $ $ \rho_{\mathrm{a}} \mathrm{T}_{\mathrm{a}}=\rho \mathrm{T} $
$ \Rightarrow $ 1.2 $ \times $ 300 = $\rho $ $ \times $ 360
$\Rightarrow \rho=1 \mathrm{~kg} / \mathrm{m}^3$
Buoyant force applied on the hot air = $\rho_{\mathrm{a}}$Vg (Upward direction)
Weight of the hot air = $\rho $Vg (Downward direction)
$ \therefore $ Net force on the hot air = $\rho_{\mathrm{a}}$Vg - $\rho $Vg
Let acceleration of the hot air in the upward direction = $a$
and mass of the hot air = $\rho $V
$ \therefore $ $\rho $V$a$ = $\rho_{\mathrm{a}}$Vg - $\rho $Vg
$ \Rightarrow $ $a$ = ${{{\rho _a}Vg - \rho Vg} \over {\rho V}}$
= ${{{\rho _a}g - \rho g} \over \rho }$
= ${{1.2 \times 10 - 1 \times 10} \over 1}$
= 2 m/s2
$ \therefore $ Velocity(v) of the hot air when exiting the chimney using formula ${v^2} = {u^2} + 2ah$,
${v^2} = 0 + 2 \times 2 \times 9$
$ \Rightarrow $ v = 6 m/s
Mass flow rate = ${{dm} \over {dt}}$ = $\rho $Av
= $ \rho \times \frac{\pi \mathrm{d}^2}{4} \times\mathrm{v}=1 \times \frac{\pi}{4} \times 10^{-2} \times 6 $
= ${{471} \over {10000}}$ kg/s = ${{471} \over {10000}} \times 1000$ gm/s = 47.1
Explanation:
For a gas R and M are constant. So, $\rho T$ = Constant (for constant pressure).
The density of hot air inside the furnace is = $\rho $
The air gets heated inside the furnace at constant pressure Pa.
$ \therefore $ $ \rho_{\mathrm{a}} \mathrm{T}_{\mathrm{a}}=\rho \mathrm{T} $
$ \Rightarrow $ 1.2 $ \times $ 300 = $\rho $ $ \times $ 360
$\Rightarrow \rho=1 \mathrm{~kg} / \mathrm{m}^3$
After chimney is closed,
Pressure at the bottom surface, P1 = Pa - $\rho $gh = Pa - (1)(10)(9)
Pressure at the bottom surface, P2 = Pa - $\rho $ag(h + H) = Pa - (1.2)(10)(9 + 1)
$ \therefore $ Pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap
= P1 - P2
= Pa - (1)(10)(9) - Pa + (1.2)(10)(9 + 1)
= 120 - 90 = 30
Explanation:
For a monatomic ideal gas, the heat capacity at constant volume, $C_{v1}$, is $\left(\frac{3}{2}\right)R$. This comes from the degrees of freedom for a monatomic gas, which are three (x, y, and z motions).
For a diatomic ideal gas, the heat capacity at constant volume, $C_{v2}$, is $\left(\frac{5}{2}\right)R$. This comes from the degrees of freedom for a diatomic gas, which are five (x, y, and z motions and rotation about two axes).
Next, we find the average heat capacity at constant volume for the gas mixture, $C_{v_{\text{mix}}}$, which is a weighted average based on the number of moles of each gas :
$C_{v_{\text{mix}}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2 \times \left(\frac{3}{2}\right)R + 1 \times \left(\frac{5}{2}\right)R}{2 + 1} = \frac{11R}{6}$
The change in internal energy $\Delta U$ for a given number of moles $n$ and change in temperature $\Delta T$ is given by :
$\Delta U = n C_{v} \Delta T$
Given that the work done by the system at constant pressure is :
$W = n R \Delta T$
We can replace $\Delta T = \frac{W}{nR}$ in the equation for $\Delta U$ to get :
$\Delta U = n C_{v_{\text{mix}}} \times \frac{W}{n R} = C_{v_{\text{mix}}} \times \frac{W}{ R} = \frac{11R}{6} \times \frac{66}{ R} = 121 \, \text{Joule}$
So the change in internal energy is indeed 121 Joule.
[Given: Wien's constant as $2.9 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ and $\frac{h c}{e}=1.24 \times 10^{-6} \mathrm{~V}-\mathrm{m}$ ]
| List - I | List - II |
|---|---|
| (P) $2000 \mathrm{~K}$ | (1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \mathrm{eV}$. |
| (Q) $3000 \mathrm{~K}$ | (2) The radiation at peak wavelength is visible to human eye. |
| (R) $5000 \mathrm{~K}$ | (3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction. |
| (S) $10000 \mathrm{~K}$ | (4) The power emitted per unit area is $1 / 16$ of that emitted by a blackbody at temperature $6000 \mathrm{~K}$. |
| (5) The radiation at peak emission wavelength can be used to image human bones. |
Steam at $100^{\circ} \mathrm{C}$ is added to 150 g water to increase its temperature from $20^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The total mass of the water at $40^{\circ} \mathrm{C}$ is (specific heat capacity of water $=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat of steam $\left.=540 \mathrm{cal} \mathrm{g}^{-1}\right)$
155 g
150 g
145 g
5 g
A blacksmith fixes circular iron frame on the wooden wheel of a bullock cart. The diameter of wooden wheel and circular iron frame are 5.012 m and 5 m respectively at $27^{\circ} \mathrm{C}$. The temperature (in ${ }^{\circ} \mathrm{C}$ ) to which iron ring must be heated so as to fit the wooden wheel is
(Coefficient of linear expansion of iron $\left.=1.2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)$
200
227
254
300
Two moles of triatomic gas $\left(\gamma=\frac{4}{3}\right)$ at temperature $327^{\circ} \mathrm{C}$ expands adiabatically such that its volume becomes 8 times its initial volume. Later the temperature of the gas is doubled in an isochoric process. The total work done in the two processes is
(Where, R is universal gas constant)
$900 R$
$1800 R$
$1200 R$
$300 R$
If the temperature of a gas is increased from $27^{\circ} \mathrm{C}$ to $159^{\circ} \mathrm{C}$, then the percentage increase in the rms speed of the gas molecules is
5
10
15
20
The Fahrenheit and Kelvin scales of temperature will have the same reading at a temperature of
$-40^{\circ} \mathrm{F}$
$313^{\circ} \mathrm{F}$
$574.6^{\circ} \mathrm{F}$
$732.7^{\circ} \mathrm{F}$
If the ratio of densities of two substances is $5: 6$ and the ratio of their specific heat capacities is $3: 5$, then the ratio of heat energies required per unit volume so that the two substances can have same temperature rise is
$1: 1$
$1: 4$
$1: 2$
$1: 3$
In a process, the work done by the system is equal to the decrease in its internal energy. The process that the system undergoes is
isothermal process
adiabatic process
isobaric process
isochoric process
N molecules each of mass $m$ of gas $A$ and 2 N molecules each of mass 2 m of gas $B$ are contained in a vessel which is maintained at a temperature $T$. The mean square velocity of the molecules of gas $B$ is denoted by $v_2^2$ and the mean square of the $x$-component velocity of the molecules of gas $A$ is denoted by $v_1^2$, then $v_1 / v_2$ is
2
1
$2 / 3$
$\sqrt{2 / 3}$