Heat and Thermodynamics
Ten moles of an ideal monoatomic gas, initially in state $\boldsymbol{a}$ at atmospheric pressure and temperature $T_a=27^{\circ} \mathrm{C}$, is enclosed in a metal cylinder of volume $V_0$ fitted with a frictionless piston. The gas is suddenly compressed to state $\boldsymbol{b}$ with volume $V_0 / 3$. Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature $11^{\circ} \mathrm{C}$ until the gas reaches the temperature of the water bath, which is denoted as state $\boldsymbol{c}$. Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state $\boldsymbol{f}$. If $R$ is universal gas constant, then the correct option(s) is/are :
[Given: $9^{1 / 3}=2.08$ ]
The schematic P-V diagram of the processes described above is :
The change in internal energy in going from state a to b is $4860R$.
The net change in the internal energy in the whole process is $-240R$.
The pressure and temperature of the state b are $2.08$ times the atmospheric pressure and $624\,K$, respectively.
A quasi-static cycle of a monoatomic ideal gas contains an isothermal process $(ab)$, followed by an isochoric process $(bc)$ and an adiabatic process $(ca)$ as shown in the figure. The volumes of the gas are $V_1$ and $V_2$ at $a$ and $b$, respectively. If the cycle has heat input $Q_{\mathrm{in}}$ and output $Q_{\mathrm{out}}$, then the efficiency of the cycle is defined as $\eta = \frac{Q_{\mathrm{in}} - Q_{\mathrm{out}}}{Q_{\mathrm{in}}}.$ The correct statement(s) is/are:
[Given: $\ln 2 \approx 0.7$]
If $\dfrac{V_2}{V_1} = 8$, the heat released in the process bc is smaller than the heat absorbed in the process ab.
For a given value of $V_2/V_1$, $\eta$ does not depend on the temperature of the isothermal process.
If $V_2/V_1 = 8$, then the temperature of the gas at a is $4$ times the temperature of the gas at c.
If $V_2/V_1 = 8$, then the pressure of the gas at a is $4$ times the pressure of the gas at b.
As shown in the figure, five Carnot engines, each with efficiency $\eta$ and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engine is completely absorbed by the next engine. Consider $Q_0$ to be the amount of heat absorbed per cycle by the first engine and $W$ as the amount of total work done by all the engines per cycle, then the net efficiency of the system is found to be
$\eta_{\mathrm{net}} = \frac{W}{Q_0} = \frac{211}{243}.$
The value of $\eta$ is:
Explanation:
Each engine has the exact same individual thermal efficiency, $\eta$.
The efficiency of any heat engine is defined as the ratio of work done $\left(\mathrm{W}_{\mathrm{i}}\right)$ to the heat absorbed $\left(\mathrm{Q}_{\mathrm{in}, \mathrm{i}}\right)$ :
$ \eta=\frac{W_i}{Q_{i n, i}}=1-\frac{Q_{o u t, i}}{Q_{i n, i}} $
So, the ratio of heat rejected to heat absorbed for any single engine in the series is :
$ \frac{\mathrm{Q}_{o u t, \mathrm{i}}}{\mathrm{Q}_{\mathrm{in}, \mathrm{i}}}=1-\eta $
For engine $\mathrm{E}_1$
Absorbed heat is $Q_0$ from the first reservoir.
Rejected heat is $Q_1=Q_0(1-\eta)$
Work done is $\mathrm{W}_1=\eta \mathrm{Q}_0$.
For engine $\mathrm{E}_2$ :
Absorbs heat rejected by $E_1$, so $Q_{\text {absorbed }, 2}=Q_1=Q_0(1-\eta)$.
Rejected heat is $Q_2=Q_1(1-\eta)=Q_0(1-\eta)^2$
Work done is $\mathrm{W}_2=\eta \mathrm{Q}_1=\eta \mathrm{Q}_0(1-\eta)$
For engine $\mathrm{E}_3$ :
Absorbed heat is $Q_2$.
Rejected heat is $Q_3=Q_2(1-\eta)=Q_0(1-\eta)^3$.
Work done $\mathrm{W}_3=\eta \mathrm{Q}_2=\eta \mathrm{Q}_0(1-\eta)^2$
For engine $\mathrm{E}_4$ :
Absorbed heat is $Q_3$.
Rejects heat is $Q_4=Q_3(1-\eta)=Q_0(1-\eta)^4$.
Work done is $\mathrm{W}_4=\eta \mathrm{Q}_3=\eta \mathrm{Q}_0(1-\eta)^3$.
For engine $\mathrm{E}_5$ :
Absorbed heat is $Q_4$.
Rejected heat is $Q_5=Q_4(1-\eta)=Q_0(1-\eta)^5$.
Work done is $\mathrm{W}_5=\eta \mathrm{Q}_4=\eta \mathrm{Q}_0(1-\eta)^4$
The total work W performed by all 5 engines combined per cycle is the sum of their individual works:
$ \mathrm{W}=\mathrm{W}_1+\mathrm{W}_2+\mathrm{W}_3+\mathrm{W}_4+\mathrm{W}_5 $
By the principle of energy conservation across the entire system, the total work done must equal the initial heat input to the first engine minus the final heat rejected by the last engine :
$ \mathrm{W}=\mathrm{Q}_0-\mathrm{Q}_5 $
$ W=Q_0-Q_0(1-\eta)^5=Q_0\left[1-(1-\eta)^5\right] $
The net efficiency $\eta_{\text {net }}$ of the combined system is given by:
$ \eta_{\text {net }}=\frac{W}{Q_0}=1-(1-\eta)^5 $
It is given that $\eta_{\text {net }}=\frac{211}{243}$.
$ 1-(1-\eta)^5=\frac{211}{243} $
$ \begin{gathered} (1-\eta)^5=1-\frac{211}{243} \\ (1-\eta)^5=\frac{243-211}{243} \\ (1-\eta)^5=\frac{32}{243} \\ 1-\eta=\left(\frac{32}{243}\right)^{1 / 5} \\ 1-\eta=\frac{2}{3} \\ \eta=1-\frac{2}{3}=\frac{1}{3} \end{gathered} $
Therefore, the value of $\eta$ is $\frac{1}{3}$
As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition ($P_1$) and a freely movable but thermally insulated piston ($P_2$). The partition $P_1$ with thermal conductivity $K$, cross sectional area $A$ and width $x$ divides the container into two sections, $S_1$ and $S_2$, each containing one mole of a monoatomic gas. The piston $P_2$ moves freely such that the gas in $S_2$ is always at the atmospheric pressure. Initially, the difference between the temperatures of $S_1$ and $S_2$ is $\Delta T_0$. The time it takes for the temperature difference to become $\frac{\Delta T_0}{2}$ is $nxR/KA$, where $R$ is the universal gas constant. The value of $n$ is:
[ Given: $ln 2 \approx 0.7$ ]
Explanation:
The two compartments, $S_1$ and $S_2$, separated by a rigid, conducting partition $P_1$.
Section $S_1$ is bounded by the outer insulated container walls and the immovable, rigid partition $P_1$.
Because the boundaries cannot move, its volume remains completely constant $\mathrm{V}=$ constant. Therefore, the gas in $\mathrm{S}_1$ undergoes an isochoric (constant volume) process.
Section $\mathrm{S}_2$ is bounded on one side by a freely movable, frictionless, insulated piston $\mathrm{P}_2$.
Since it moves freely to balance the external atmospheric pressure, the pressure of the gas inside $S_2$ remains constant $\mathrm{P}=$ constant $=\mathrm{P}_{\mathrm{atm}}$.
Therefore, the gas in $\mathrm{S}_2$ undergoes an isobaric (constant pressure) process.
Both compartments contain $\mathrm{n}_{\mathrm{m}}=1$ mole of an ideal monoatomic gas.
For a monoatomic ideal gas, the molar heat capacity at constant volume $\left(\mathrm{C}_{\mathrm{v}}\right)$ is $\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}$
The molar heat capacity at constant pressure $(C_p)$ is given by $C_p = C_v + R$
$C_p = \frac{3}{2}R + R = \frac{5}{2}R$
Let $T_1$ be the instantaneous temperature of the gas in $S_1$ and $T_2$ be the instantaneous temperature of the gas in $S_2$ at any time $t$.
Let us assume that $S_1$ is initially at a higher temperature than $S_2$, so $T_1 > T_2$.
As time progresses, heat $dQ$ flows across the conducting partition $P_1$ from the hotter section $S_1$ to the cooler section $S_2$.
For section $S_1$ the gas undergoes isochoric cooling.
The heat lost by $S_1$ causes its temperature to decrease by $dT_1$:
$dQ = -n_m C_v\, dT_1 = -(1)\left(\frac{3}{2}R\right)dT_1 \rightarrow dT_1 = -\frac{2}{3R}dQ$
For section $S_2$ where the gas undergoes isobaric heating.
The same amount of heat $dQ$ enters $S_2$, causing its temperature to increase by $dT_2$:
$dQ = n_m C_p\, dT_2 = (1)\left(\frac{5}{2}R\right)dT_2 \rightarrow dT_2 = \frac{2}{5R}dQ$
$d(\Delta T) = dT_1 - dT_2$
$d(\Delta T) = -\frac{2}{3R}dQ - \frac{2}{5R}dQ = -\left(\frac{2}{3R} + \frac{2}{5R}\right)dQ$
$d(\Delta T) = -\left(\frac{10 + 6}{15R}\right)dQ = -\frac{16}{15R}dQ$
According to the rate of heat flow $\frac{dQ}{dt}$ through a conducting barrier of thermal conductivity $K$, area $A$, and thickness $x$ under a temperature difference $(T_1 - T_2)$ is given by:
$\frac{dQ}{dt} = \frac{KA}{x}(T_1 - T_2) = \frac{KA}{x}\,\Delta T$
Therefore, the infinitesimal heat transferred in a time interval $dt$ is:
$dQ = \frac{KA}{x}\Delta T\, dt$
Substituting the expression for $dQ$,
$d(\Delta T) = -\frac{16}{15R}\left(\frac{KA}{x}\Delta T\right)$
$\frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR}dt$
At $\mathrm{t}=0$, the initial temperature difference is $\Delta \mathrm{T}=\Delta \mathrm{T}_0$.
At $\mathrm{t}=\tau$, the final temperature difference becomes $\Delta \mathrm{T}=\frac{\Delta \mathrm{T}_0}{2}$.
$ \int\limits_{\Delta \mathrm{T}_0}^{\Delta \mathrm{T}_0 / 2} \frac{\mathrm{~d}(\Delta \mathrm{~T})}{\Delta \mathrm{T}}=-\frac{16 \mathrm{KA}}{15 \times \mathrm{R}} \int\limits_0^\tau \mathrm{dt} $
$\Rightarrow $ $[ n\left(\Delta T\right)]^{\Delta T_0/2}_{\Delta T_0} = -\frac{16KA}{15xR}\,\tau$
$\Rightarrow $ $\ln\left(\frac{\Delta T_0/2}{\Delta T_0}\right) = -\frac{16KA}{15xR}\,\tau$
$\Rightarrow $ $\ln \frac{1}{2} = -\frac{16KA}{15xR}\,\tau$
$\Rightarrow $ $-\ln 2 = -\frac{16KA}{xR}\,\tau \Rightarrow \ln 2 = \frac{16KA}{15xR}\,\tau$
$\Rightarrow $ $\tau = \frac{15}{16}\ln(2)\frac{xR}{KA}$
Using the given approximation $\ln 2 \approx 0.7$:
$\tau = \frac{15}{16} \times (0.7) \times \frac{xR}{KA} = \frac{10.5\,xR}{16\,KA} = 0.65625\,\frac{xR}{KA}$
$\Rightarrow n \approx 0.66$
Therefore, the correct answer is 0.66.
The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4 . It extracts 150 J of heat per cycle from the hot reservoir. The work extracted from this engine is being fully used to run a heat pump which has a coefficient of performance 10 . The hot reservoir of the heat pump is at a temperature of 300 K . Which of the following statements is/are correct :
Work extracted from the Carnot engine in one cycle is 60 J.
Temperature of the cold reservoir of the Carnot engine is 600 K.
Temperature of the cold reservoir of the heat pump is 270 K.
Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.
Explanation:
$ \begin{aligned} & \mathrm{nRT}_{\mathrm{w}}=\mathrm{P}_{\mathrm{w}} \mathrm{~V}_{\mathrm{w}}=1 \mathrm{~J} \\ & \mathrm{P}_{\mathrm{W}}=\frac{1}{64} \times 10^6 \mathrm{~Pa} \end{aligned} $
For WX process
$ \begin{aligned} & P_X V_X^y=P_W V_W^y \\ & \Rightarrow P_X=P_W\left(\frac{V_W}{V_X}\right)^y \end{aligned} $
amount of heat absorbed in XY process
$ \begin{aligned} \mathrm{Q} & =\mathrm{nCP} \Delta \mathrm{~T}=\mathrm{n} \times \frac{5}{2} \mathrm{R} \times\left[\mathrm{T}_{\mathrm{Y}}-\mathrm{T}_{\mathrm{X}}\right] \quad\left[\text { For monoatomic gas } \mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2}\right] \\ \mathrm{Q} & =\frac{5}{2}\left[\mathrm{nRT}_{\mathrm{Y}}-\mathrm{nRT}_{\mathrm{X}}\right] \\ & =\frac{5}{2}\left[\mathrm{P}_{\mathrm{Y}} \mathrm{~V}_{\mathrm{Y}}-\mathrm{P}_{\mathrm{X}} \mathrm{~V}_{\mathrm{X}}\right] \\ & =\frac{5}{2} \mathrm{P}_{\mathrm{X}}\left[\mathrm{~V}_{\mathrm{Y}}-\mathrm{V}_{\mathrm{X}}\right] \quad\left[\because \mathrm{P}_{\mathrm{X}}=\mathrm{P}_{\mathrm{Y}} ; \text { Isobaric process }\right] \\ & =\frac{5}{2} \times \mathrm{P}_{\mathrm{W}} \times\left[\frac{\mathrm{V}_{\mathrm{W}}}{\mathrm{~V}_{\mathrm{X}}}\right]^{\mathrm{y}}\left[\mathrm{~V}_{\mathrm{Y}}-\mathrm{V}_{\mathrm{X}}\right] \end{aligned} $
Putting values :
Q = 1.6 Joule
Explanation:
Extension in spring
$ \begin{aligned} & \mathrm{x}=0.5 \mathrm{~L}-0.4 \mathrm{~L} \\ & =0.1 \mathrm{~L} \end{aligned} $
FBD of piston
$ \begin{aligned} & \mathrm{kx}+\mathrm{P}_2 \mathrm{~A}=\mathrm{P}_1 \mathrm{~A} \\ & \mathrm{P}_2 \mathrm{~A}=\mathrm{P}_1 \mathrm{~A}-\mathrm{kx} \\ & \mathrm{P}_2=\mathrm{P}_1-\frac{\mathrm{kL}}{\mathrm{~A}(10)} ........(i) \\ & \mathrm{P}_1 \mathrm{~V}=\mathrm{n}_1 \mathrm{RT} \\ & \mathrm{P}_2 \mathrm{~V}=\mathrm{n}_2 \mathrm{RT} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{2} \end{aligned} $
$ \begin{aligned} & \mathrm{P}_1=\frac{3}{2} \mathrm{P}_2 ........(ii) \\ & \mathrm{P}_2=\frac{3}{2} \mathrm{P}_2-\frac{\mathrm{kL}}{10 \mathrm{~A}} \\ & \frac{\mathrm{P}_2}{2}=\frac{\mathrm{kL}}{10 \mathrm{~A}} \\ & \mathrm{P}_2=\frac{\mathrm{kL}}{5 \mathrm{~A}}=\frac{\mathrm{kL}}{\mathrm{~A}} \alpha \\ & \alpha=\frac{1}{5}=0.2 \end{aligned} $
Two identical plates P and Q , radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures $\mathrm{T}_{\mathrm{P}}$ and $\mathrm{T}_{\mathrm{Q}}$, respectively, with $\mathrm{T}_{\mathrm{Q}}<\mathrm{T}_{\mathrm{P}}$, as shown in Fig. 1. The radiated power transferred per unit area from P to Q is $W_0$. Subsequently, two more plates, identical to P and Q , are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from $P$ to $Q$ (Fig. 2) in the steady state is $W_S$, then the ratio $\frac{W_0}{W_S}$ is ________.
Explanation:
Since the plates are blackbodies, they emit radiation according to the Stefan-Boltzmann law. The power emitted per unit area by a blackbody is given by.
$ P=\sigma T^4 . $
Where, $\sigma=$ Stefan - Boltzmann constant
$ \sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^2 \mathrm{k}^4 $
When two blackbodies face each other, the net power transfer per unit area from $P$ to $Q=$ power emitted by $P$ toward $Q$ - power emitted by $Q$ toward $P$
So,
$ W_0=\sigma T_p^4-\dot{\sigma} T_a^4=\sigma\left(T_p^4-T_a^4\right)\,\,\,\,\,\,\,\,\,\,\,\,...(i) $
After introducing two additional plates: assuming temperature $T_1$ and $T_2$ for these plates.
In the steady state, the power transfer per unit area through each interface (From $P$ to plate $1,1 \rightarrow 2,2 \rightarrow Q$ ) must be same, because there is no accumulation of energy in the intermediate plates.
So
$ \begin{aligned} & \sigma\left(T_p^4-T_1^4\right)=W_S \,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \sigma\left(T_1^4-T_2^4\right)=W_S \,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $
and $\sigma\left(T_2^4-T_a^4\right)=W_S\,\,\,\,...(iv)$
by adding $e q^n$ (2), (3) and (4),
$ \begin{aligned} & \sigma\left(T_p^4-T _1^4+T_1^4-T_2^4+T_2^4-T_Q^4\right)=3 W_S \\ \Rightarrow & \sigma\left(T_p^4-T_Q^4\right)=3 W_S \\ \Rightarrow & W_0=3 W_S \\ & \Rightarrow \frac{W_0}{W_S}=3 \end{aligned} $
Explanation:
Case - 1
$\begin{aligned} & P-P_0=\Delta P=\frac{4 T}{R} \\ & P=\left(P_0+\frac{4 T}{R}\right) \end{aligned}$
Case-2
$\begin{aligned} & \mathrm{P}_1-\frac{8 \mathrm{P}_0}{27}=\Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1} \\ & \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27} \end{aligned}$
Constant temperature process
$\begin{aligned} & \mathrm{PV}=\mathrm{P}_1 \mathrm{~V}_1 \\ & \left(\mathrm{P}_0+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4}{3} \pi \mathrm{R}^3=\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27}\right) \frac{4}{3} \pi \mathrm{R}_1^3 ;\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right),\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}\right) \rightarrow \text { (Neglected) } \\ & \mathrm{R}=\frac{2}{3} \mathrm{R}_1 \Rightarrow \mathrm{R}_1=\frac{3}{2} \mathrm{R} \\ & \Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}=\frac{4 \mathrm{~T}}{3 \mathrm{R}} \times 2=\frac{2}{3} \times(144)=96 \mathrm{~Pa} \end{aligned}$
The specific heat capacity of a substance is temperature dependent and is given by the formula $C=k T$, where $k$ is a constant of suitable dimensions in SI units, and $T$ is the absolute temperature. If the heat required to raise the temperature of $1 \mathrm{~kg}$ of the substance from $-73^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}$ is $n k$, the value of $n$ is ________.
[Given: $0 \mathrm{~K}=-273{ }^{\circ} \mathrm{C}$.]
Explanation:
To solve this problem, we need to integrate the heat capacity over the given temperature range because the specific heat capacity is temperature dependent. We are given that the specific heat capacity $C$ is defined as $C = kT$, where $k$ is a constant, and $T$ is the absolute temperature.
The heat required to raise the temperature, $Q$, can be found using the following integral:
$ Q = \int_{T_1}^{T_2} C \, dT $
Given $C = kT$, the integral becomes:
$ Q = \int_{T_1}^{T_2} kT \, dT $
We need to convert the given temperatures from Celsius to Kelvin. The temperatures given are:
- Initial temperature: $-73^{\circ} \mathrm{C}$
- Final temperature: $27^{\circ} \mathrm{C}$
In Kelvin, these temperatures are:
- $T_1 = -73^{\circ} \mathrm{C} + 273 = 200 \, \mathrm{K}$
- $T_2 = 27^{\circ} \mathrm{C} + 273 = 300 \, \mathrm{K}$
Now we can evaluate the integral:
$ Q = k \int_{200}^{300} T \, dT $
Integrating, we get:
$ Q = k \left[ \frac{T^2}{2} \right]_{200}^{300} $
Substituting the limits of integration:
$ Q = k \left[ \frac{300^2}{2} - \frac{200^2}{2} \right] $
Solving the values inside the brackets:
$ Q = k \left[ \frac{90000}{2} - \frac{40000}{2} \right] $
$ Q = k \left[ 45000 - 20000 \right] $
$ Q = k \cdot 25000 $
We are given that this heat is equal to $nk$:
$ nk = k \cdot 25000 $
Dividing both sides by $k$, we get:
$ n = 25000 $
Thus, the value of $n$ is 25000.
One mole of a monatomic ideal gas undergoes the cyclic process $\mathrm{J} \rightarrow \mathrm{K} \rightarrow \mathrm{L} \rightarrow \mathrm{M} \rightarrow \mathrm{J}$, as shown in the P-T diagram.
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
[ $\mathcal{R}$ is the gas constant.]
| List-I | List-II |
|---|---|
| (P) Work done in the complete cyclic process | (1) $RT_0 - 4RT_0 \ln 2$ |
| (Q) Change in the internal energy of the gas in the process JK | (2) $0$ |
| (R) Heat given to the gas in the process KL | (3) $3RT_0$ |
| (S) Change in the internal energy of the gas in the process MJ | (4) $-2RT_0 \ln 2$ |
| (5) $-3RT_0 \ln 2$ |
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. |
In the given $P-V$ diagram, a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is first compressed adiabatically from state $A$ to state $B$. Then it expands isothermally from state $B$ to state $C$. [Given: $\left(\frac{1}{3}\right)^{0.6} \simeq 0.5, \ln 2 \simeq 0.7$ ].
Which of the following statement(s) is(are) correct?
List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.
| List-I | List-II |
|---|---|
| (I) $10^{-3} \mathrm{~kg}$ of water at $100^{\circ} \mathrm{C}$ is converted to steam at the same temperature, at a pressure of $10^{5} \mathrm{~Pa}$. The volume of the system changes from $10^{-6} \mathrm{~m}^{3}$ to $10^{-3} \mathrm{~m}^{3}$ in the process. Latent heat of water $=2250\, \mathrm{~kJ} / \mathrm{kg}$. |
(P) $2 \mathrm{~kJ}$ |
| (II) $0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 \mathrm{~K}$ undergoes an isobaric expansion to volume $3 \mathrm{~V}$. Assume $R=8.0 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$. |
(Q) $7 k J$ |
| (III) One mole of a monatomic ideal gas is compressed adiabatically from volume $V=\frac{1}{3} \mathrm{~m}^{3}$ and pressure $2 \mathrm{kPa}$ to volume $\frac{V}{8}$. |
(R) $4 \mathrm{~kJ}$ |
| (IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 \mathrm{~kJ}$ of heat and undergoes isobaric expansion. |
(S) $5 \mathrm{~kJ}$ |
| (T) $3 \mathrm{~kJ}$ |
Which one of the following options is correct?
The value of X is _______________.
Explanation:
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ Vf = 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
The value of Y is _______________.
Explanation:
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ Vf = 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
For isothermal process,
${p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} \times {{3.3} \over 3}$
pf = 1.1 $\times$ 105
pf $-$ pi = 1.1 $\times$ 105 $-$ 105
= 0.1 $\times$ 105 = 10 $\times$ 103 Pa $\Rightarrow$ Y = 10
Explanation:
Ti = 200 K, e = 1
$ - Ms{{dT} \over {dt}} = {{dQ} \over {dt}} = \sigma eA{T^4}$
$ - {{dT} \over {dt}} = {{\sigma A{T^4}} \over {Ms}}$
${{\sigma A} \over {Ms}}\int\limits_{{t_i}}^{{t_f}} {dt = - \int\limits_{{T_i}}^{{T_f}} {{{dT} \over {{T^4}}}} } $
${{\sigma A} \over {Ms}}({t_f} - {t_i}) = {1 \over 3}\left( {{1 \over {T_f^3}} - {1 \over {T_i^3}}} \right)$
${{{{\sigma A} \over {Ms}}({t_1} - 0) = {1 \over 3}\left( {{1 \over {{{(100)}^3}}} - {1 \over {{{(200)}^3}}}} \right)} \over {{{\sigma A} \over {Ms}}({t_2} - 0) = {1 \over 3}\left( {{1 \over {{{(50)}^3}}} - {1 \over {{{(200)}^3}}}} \right)}}$
${{{t_1}} \over {{t_2}}} = {{{{{{(200)}^3} - {{(100)}^3}} \over {{{(100)}^3}{{(200)}^3}}}} \over {{{{{(200)}^3} - {{(50)}^3}} \over {{{(50)}^3}{{(200)}^3}}}}}$
$\therefore$ ${{{t_2}} \over {{t_1}}} = {9 \over 1}$
(Take Stefan-Boltzmann constant = 5.67 $ \times $ 10−8 Wm−2K−4 , Wien’s displacement constant = 2.90 $ \times $ 10−3 m-K, Planck’s constant = 6.63 $ \times $ 10−34 Js, speed of light in vacuum = 3.00 $ \times $ 108 ms−1)
in the range 3.15 $ \times $ 10−8 W to 3.25 $ \times $ 10−8 W
(take the acceleration due to gravity = 10 ms−2 and the universal gas constant = 8.3 J mol−1K−1).
Explanation:
Volumes of two compartments are
$ V_1=(4+x) A \text { and } V_2=(4-x) A $
At equilibrium,
$ F_2=F_1+m g $
$ \begin{aligned} & P_2 A=P_1 A+m g \\\\ & P_2=P_1+\frac{m g}{A} \end{aligned} $
From $P V=n R T$, we get $P=\frac{n R T}{V}$
$ \begin{aligned} & \frac{n R T}{V_2}=\frac{n R T}{V_1}+\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{1}{V_2}-\frac{1}{V_1}\right]=\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{1}{A(4-x)}-\frac{1}{4(4+x)}\right]=\frac{m g}{A} \\\\ \Rightarrow & \frac{n R T}{A}\left[\frac{(4+x)-(4-x)}{(4-x)(4+x)}\right]=\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{2 x}{\left(16-x^2\right)}\right]=m g \\\\ \Rightarrow & 0.1 \times 8.3 \times 300\left[\frac{2 x}{\left(16-x^2\right)}\right]=8.3 \times 10 \\\\ \Rightarrow & \frac{6 x}{16-x^2}=1 \Rightarrow 16-x^2=6 x \\\\ \Rightarrow & x^2+6 x-16=0 \\\\ \Rightarrow & x=\frac{-6 \pm \sqrt{36+64}}{2} \\\\ & x=\frac{-6 \pm \sqrt{100}}{2}=\frac{-6 \pm 10}{2} \\\\ & x=\frac{10-6}{2} \text { or } \frac{-10-6}{2}=-8 \text { or } 2 \end{aligned} $
Neglecting negative $\operatorname{sign} x=2$
Partition from top $=4+2=6 \mathrm{~m}$
Explanation:
$ \therefore $ Wexternal + Wwater + Wgas = 0 ....(i)
where Wexternal = work done by external agent
Wwater = work done by water
Wgas = work done by gas
For adiabatic process, ${P_1}V_1^\gamma = {P_2}V_2^\gamma $
Here, ${P_0}{\left[ {{4 \over 3}\pi {R^3}} \right]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]^{{{41} \over {30}}}}$
$ \Rightarrow $ P = ${P_0}{\left[ {{R \over {R - a}}} \right]^{{{41} \over {10}}}}$
Now, Wgas = ${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3} - P \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} \right]}^{{{41} \over {10}}}} \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} \right)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]} \over {{{11} \over {30}}}}$
= ${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} \right)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]}$
= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} \right)}^{ - {{11} \over {10}}}}} \right]$
= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} \right)\left( { - {{11} \over {10}} - 1} \right)} \over 2}{{{a^2}} \over {{R^2}}}} \right]$
= -4P0$\pi $R2$a$ - ${{40 \times 21} \over {100 \times 2}}{P_0}\pi R{a^2}$
= -4P0$\pi $R2$a$ - $4.2{P_0}\pi R{a^2}$
Wwater = P0dV
= P0${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} \right)}^3}} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} \right)} \right]\left[ {{R^2} + R\left( {R - a} \right) + {{\left( {R - a} \right)}^2}} \right]} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left[ a \right]\left[ {3{R^2} - 3Ra + {a^2}} \right]} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} \right)} \right]}$
= ${4{P_0}\pi \left[ {{R^2}a - R{a^2}} \right]}$ [ignore ${{a^3}}$ term as no ${{a^3}}$ term in the question]
$ \therefore $ Wgas + Wwater = -$4{P_0}\pi R{a^2}$ - $4.2{P_0}\pi R{a^2}$
= $ - 4{P_0}\pi R{a^2}\left[ {1 + 1.05} \right]$
= $ - 4{P_0}\pi R{a^2}\left[ {2.05} \right]$
From (i), Wexternal = - (Wgas + Wwater)
= $4{P_0}\pi R{a^2}\left[ {2.05} \right]$
$ \therefore $ X = 2.05
Explanation:
For small temperature change,
${{dQ} \over {dt}} = e\sigma A{T^3}\Delta T$ .... (i)
${{mCdT} \over {dt}} = e\sigma A{T^3}\Delta T $
$\Rightarrow {{dT} \over {dt}} = {{e\sigma A{T^3}} \over {mC}}\Delta T$
${{e\sigma A{T^3}} \over {mC}}$ $\to$ constant for Newton law of cooling
$ \therefore $ ${{e\sigma A{T^3}} \over {mC}} = 0.001 $
$\Rightarrow e\sigma A{T^3} = mC \times 0.001 = 1 \times 4200 \times 0.001$
$e\sigma A{T^3} = 4.2$ .... (ii)
${{dQ} \over {dt}} = 700 \times 0.05 = 35$ W ..... (iii)
Putting the value of Eqs. (ii) and (iii) in Eq. (i), we get
$35 = 4.2\Delta T \Rightarrow {{35} \over {4.2}} = \Delta T \Rightarrow \Delta T = 8.33$
Explanation:
${{{p_1}} \over 4}{(4{V_1})^{5/3}} = {p_2}{(32{V_1})^{5/3}}$
${p_2} = {{{p_1}} \over 4}{\left( {{1 \over 8}} \right)^{5/3}} = {{{p_1}} \over {128}}$
${W_{adi}} = {{{p_1}{V_1} - {p_2}{V_2}} \over {\gamma - 1}}$
$ = {{{p_1}{V_1} - {{{p_1}} \over {128}}(32{V_1})} \over {{5 \over 3} - 1}}$
$ = {{{p_1}{V_1}(3/4)} \over {2/3}} = {9 \over 8}{p_1}{V_1}$
${W_{iso}} = {p_1}{V_1}\ln \left( {{{4{V_1}} \over {{V_1}}}} \right) = 2{p_1}{V_1}\ln 2$
$ \therefore $ ${{{W_{iso}}} \over {{W_{adi}}}} = {{16} \over 9}\ln 2$
$ \Rightarrow f = {{16} \over 9} = 1.7778 \approx 1.78$
(Given, 21.2 = 2.3; 23.2 = 9.2; R is a gas constant)
Explanation:
The degrees of freedom of a monatomic gas is f = 3. Its molar specific heats are ${C_v} = {f \over 2}RT$ and ${C_p} = {C_v} + R = {{(f + 2)} \over 2}RT$. The ratio of specific heats is given by
$\gamma = {C_p}/{C_v} = (f + 2)/f = 5/3$.
In an adiabatic process, $TV_{}^{\gamma - 1}$ = constant. Hence, ${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$, which gives
${T_2} = {T_1}{({V_1}/{V_2})^{\gamma - 1}} = 100{(1/8)^{5/3 - 1}} = 25\,K$,
where we used T1 = 100 K and V2 = 8V1. The change in internal energy of one mole of the ideal gas is
$\Delta U = {U_2} - {U_1} = {f \over 2}R{T_2} - {f \over 2}R{T_1} = {f \over 2}R({T_2} - {T_1})$
$ = (3/2)(8)(25 - 100) = - 900\,J$.
Thus, the decrease in internal energy of the gas is 900 J.
| LIST - I | LIST - II | ||
|---|---|---|---|
| P. | In process I | 1. | Work done by the gas is zero |
| Q. | In process II | 2. | Temperature of the gas remains unchanged |
| R. | In process III | 3. | No heat is exchanged between the gas and its surroundings |
| S. | In process IV | 4. | Work done by the gas is 6P0V0 |
$\Delta U = \Delta Q - P\Delta V$?
Explanation:
Therefore, ${{{p_1}} \over {{p_0}}} = 2$
According to Stefan's law, p $ \propto $ T2
$ \Rightarrow {{{p_2}} \over {{p_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4} = {\left( {{{2767 + 273} \over {487 + 273}}} \right)^4} = {4^4}$
${{{p_2}} \over {{p_1}}} = {{{p_2}} \over {2{p_0}}} = {4^4} \Rightarrow {{{p_2}} \over {{p_0}}} = 2 \times {4^4}$
${\log _2}{{{p_2}} \over {{p_0}}} = {\log _2}[2 \times {4^4}]$
$ = {\log _2}2 + {\log _2}{4^4}$
$ = 1 + {\log _2}{2^8} = 1 + 8 = 9$
The temperature of water fed into the device cannot exceed 30°C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is :
(Specific heat of water is 4.2 kJ kg−1 K−1 and the density of water is 1000 kg m−3)
Ignoring the friction between the piston and the cylinder, the correct statements is/are
Explanation:
Power, $P = (\sigma {T^4}A) = \sigma {T^4}(4\pi {R^2})$
or, $P \propto {T^4}{R^2}$ ..... (i)
According to Wien's law,
$\lambda \propto {1 \over T}$
($\lambda$ is the wavelength at which peak occurs)
$\therefore$ Eq. (i) will become,
$P \propto {{{R^2}} \over {{\lambda ^4}}}$
or, $\lambda \propto {\left[ {{{{R^2}} \over P}} \right]^{1/4}}$
$ \Rightarrow {{{\lambda _A}} \over {{\lambda _B}}} = {\left[ {{{{R_A}} \over {{R_B}}}} \right]^{1/2}}{\left[ {{{{P_B}} \over {{P_A}}}} \right]^{1/4}}$
$ = {[400]^{1/2}}{\left[ {{1 \over {{{10}^4}}}} \right]^{1/4}} = 2$
Explanation:
The first law of thermodynamics for the process iaf gives
${Q_{iaf}} = {U_{iaf}} + {W_{iaf}} = ({U_f} - {U_i}) + ({W_{ia}} + {W_{af}})$ ..... (1)
Substitute Qiaf = 500 J, Ui = 100 J, Wia = 0 (constant volume), and Waf = 200 J in equation (1) to get Uf = 400 J.
In the process ib,
${Q_{ib}} = {U_{ib}} + {W_{ib}} = ({U_b} - {U_i}) + {W_{ib}}$ ..... (2)
Substitute Ub = 200 J, Ui = 100 J, and Wib = 50 J in equation (2) to get Qib = 150 J.
In the process bf,
${Q_{bf}} = {U_{bf}} + {W_{bf}} = ({U_f} - {U_b}) + {W_{bf}}$ ..... (3)
Substitute Uf = 400 J, Ub = 200 J and Wbf = 100 J in equation (3) to get Qbf = 300 J. Thus, Qbf/Qib = 300/150 = 2.