In-Class Practice Sheet: Thermodynamics (JEE Level)

Solution:
According to the first law of thermodynamics: $ \Delta Q = \Delta U + \Delta W $
Heat absorbed, $\Delta Q = +400 \text{ J}$
Work done by strictly, $\Delta W = +100 \text{ J}$
$ 400 = \Delta U + 100 \implies \Delta U = 300 \text{ J} $
The internal energy increases by 300 J.

Solution:
$\Delta U$ is a state function and is strictly path-independent. It depends only on states A and B.
$\Delta Q$ and $\Delta W$ are path variables. Therefore, heat absorbed ($\Delta Q$) is path-dependent.

Solution:
Given $P = kV$. From ideal gas law, $PV = RT \implies (kV)V = RT \implies kV^2 = RT$.
Differentiating: $k(2V dV) = R dT \implies P dV = \frac{R}{2} dT$
Work done $dW = P dV = \frac{R}{2} dT$.
Change in internal energy $dU = n C_v dT = \frac{3R}{2} dT$.
Heat supplied $dQ = dU + dW = \frac{3R}{2} dT + \frac{R}{2} dT = 2R dT$
Total heat supplied = $2R\Delta T$.

Solution:
Work done at constant external pressure: $W = P_{ext} \Delta V$
$W = (2 \times 1.01 \times 10^5 \text{ Pa}) \times (20 - 10) \times 10^{-3} \text{ m}^3$
$W = 2.02 \times 10^5 \times 10 \times 10^{-3} = 2020 \text{ J}$.

Solution:
Let initial pressure be $P_0 = P_{atm}$. $P_0 = \frac{RT_0}{V_0}$.
Let the piston move up by $x$. Final volume $V = V_0 + Ax$.
The gas pressure $P$ must balance the atmospheric pressure and the spring force: $P A = P_{atm} A + kx \implies P = P_{atm} + \frac{kx}{A}$.
So $P$ increases linearly with $V$: $P = P_0 + \frac{k}{A^2}(V - V_0)$.
The process is a straight line on the P-V diagram.
Work done $W = \text{Area under P-V curve} = \frac{1}{2}(P_0 + P)(V - V_0)$.
Substitute $P - P_0 = \frac{kx}{A}$ and $V - V_0 = Ax$:
$W = P_{atm}(Ax) + \frac{1}{2}kx^2$.
To find $x$, use the ideal gas law at the final state $T_1$: $(P_{atm} + \frac{kx}{A})(V_0 + Ax) = RT_1$.
Solving this quadratic equation gives $x$, and substituting it into the work equation yields the final exact answer.

Solution:
Degrees of freedom for a diatomic gas (translational + rotational) $f = 3 + 2 = 5$.
$C_v = \frac{f}{2}R = \frac{5}{2}R$
$C_p = C_v + R = \frac{7}{2}R$
$\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = 1.4$

Solution:
$O_2$ is diatomic ($f_1=5, n_1=2$). $C_{v1} = \frac{5}{2}R$
$Ar$ is monoatomic ($f_2=3, n_2=4$). $C_{v2} = \frac{3}{2}R$
Equivalent $C_v$ of mixture: $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$
$C_{v,mix} = \frac{2(\frac{5}{2}R) + 4(\frac{3}{2}R)}{2 + 4} = \frac{5R + 6R}{6} = \frac{11R}{6}$

Solution:
Mayer's relation: $C_p - C_v = R \implies 5 - C_v = 2 \implies C_v = 3 \text{ cal mol}^{-1}\text{K}^{-1}$
Since $C_v = \frac{3}{2}R$, using $R$ as 2, $\frac{3}{2}(2) = 3$. This perfectly matches.
Since $C_v = \frac{3}{2}R$, the gas has 3 degrees of freedom. Therefore, it is a monoatomic gas.

Solution:
In an isothermal process, $\Delta U = 0$, so $\Delta Q = \Delta W$.
$W = nRT \ln\left(\frac{V_2}{V_1}\right) = 2 \times 8.31 \times 300 \times \ln(2)$
$W = 2 \times 8.31 \times 300 \times 0.693 \approx 3455.5 \text{ J}$
Heat absorbed = 3455.5 J.

Solution:
1. The walls of the container must be perfectly conducting to allow heat exchange.
2. The process must be carried out infinitesimally slowly (quasi-static) so that the system constantly remains in thermal equilibrium with the surroundings.

Solution:
Since $PV = \text{const}$, $P_1(V) = P_2(2V) \implies P_2 = P_1 / 2$. Pressure becomes half.
Isothermal Bulk Modulus ($K_{iso}$): $K = -V \frac{dP}{dV}$.
From $PV = c \implies P dV + V dP = 0 \implies -V\frac{dP}{dV} = P$.
Therefore, Isothermal Bulk Modulus is exactly equal to the Pressure of the gas, $K_{iso} = P$.

Solution:
Equation of state: $T V^{\gamma - 1} = \text{const}$
$T_1 V_1^{1.4 - 1} = T_2 V_2^{1.4 - 1} \implies T_1 V_1^{0.4} = T_2 \left(\frac{V_1}{32}\right)^{0.4}$
$\frac{T_2}{T_1} = (32)^{0.4} = (2^5)^{2/5} = 2^2 = 4$.

Solution:
Isothermal slope: $PV = c \implies P dV + V dP = 0 \implies \left(\frac{dP}{dV}\right)_{iso} = -\frac{P}{V}$
Adiabatic slope: $PV^\gamma = k \implies \gamma P V^{\gamma-1} dV + V^\gamma dP = 0$
Divide by $V^{\gamma-1}$: $\gamma P dV + V dP = 0 \implies \left(\frac{dP}{dV}\right)_{adia} = -\gamma \frac{P}{V}$
Hence, $\left(\frac{dP}{dV}\right)_{adia} = \gamma \left(\frac{dP}{dV}\right)_{iso}$. Since $\gamma > 1$, it is strictly steeper.

Solution:
For an adiabatic process, $\Delta Q = 0$.
From 1st Law: $0 = \Delta U + \Delta W \implies \Delta W = -\Delta U$
$\Delta W = -(-50 \text{ J}) = +50 \text{ J}$. Work done by the gas is 50 J.

Solution:
This is an Adiabatic Free Expansion.
Since it expands into vacuum (absence of opposing atmospheric pressure), the external pressure is zero, so Work Done $W = \int P_{ext} dV = 0$.
Since the cylinder is non-conducting, Heat Exchange $Q = 0$.
From 1st Law, $\Delta U = Q - W = 0 - 0 = 0$.
Since $\Delta U = 0$ for an ideal gas depends solely on temperature, $\Delta T = 0$. The final temperature is still $T_0$.
This is a highly irreversible natural process (spontaneous expansion). Even though the initial and final temperatures are identical, the entropy of the universe increases. The change in entropy corresponds to an equivalent reversible isothermal process expanding from $V_0$ to $2V_0$: $\Delta S = nR \ln(2V_0 / V_0) = R \ln 2$.

Solution:
From 1st Law: $dQ = dU + dW \implies n C dT = n C_v dT + P dV$
Differentiating $PV^n = c$: $V^n dP + n P V^{n-1} dV = 0 \implies dP = -\frac{nP}{V} dV$
Also $PV = nRT \implies P dV + V dP = nR dT$
Substitute $dP$: $P dV - n P dV = nR dT \implies P dV(1 - n) = nR dT$
So, $P dV = \frac{nR dT}{1 - n}$
Substituting this back into 1st Law step 1:
$n C dT = n C_v dT + \frac{nR dT}{1 - n}$
$C = C_v + \frac{R}{1 - n}$

Solution:
$P \propto V \implies PV^{-1} = \text{const}$. Here, $n = -1$.
Using the polytropic specific heat formula: $C = C_v + \frac{R}{1 - n}$
$C = \frac{3}{2}R + \frac{R}{1 - (-1)} = \frac{3}{2}R + \frac{R}{2} = 2R$.
Thus, the molar heat capacity is $2R$.

Solution:
Net work done is the area of the rectangle ABCD.
Base = $2V_0 - V_0 = V_0$. Height = $2P_0 - P_0 = P_0$. Area = $P_0 V_0$.
Since sequence is A->B->C->D->A is counter-clockwise? Wait, A(V0,P0) -> B(2V0,P0) -> C(2V0,2P0) -> D(V0,2P0) -> A.
This path traverses the area in a counter-clockwise direction on a standard P-Y, V-X axes.
Net Work for Counter-Clockwise cycle = Negative Area = $-P_0 V_0$.

Solution:
For any cyclic process, initial state = final state, so $\Delta U = 0$.
From 1st Law, $\Delta Q = \Delta W_{net}$.
The work done is the area of the circle. If axes scales are properly aligned: $\Delta W = \pi r^2$. (Assuming clockwise).
Therefore, net heat absorbed $\Delta Q = \pi r^2$.

Solution:
From the ideal gas law for 1 mole, $T = \frac{PV}{R}$.
Substitute $P$ into this relation: $T = \frac{(P_0 - \alpha V^2)V}{R} = \frac{P_0 V - \alpha V^3}{R}$.
To find the maximum possible temperature, take the derivative with respect to $V$ and set it to zero:
$\frac{dT}{dV} = \frac{P_0 - 3\alpha V^2}{R} = 0 \implies V = \sqrt{\frac{P_0}{3\alpha}}$.
Now plug this optimized Volume $V$ back into the temperature equation to find the maximum:
$T_{max} = \frac{P_0 \sqrt{P_0 / 3\alpha} - \alpha (P_0 / 3\alpha)^{3/2}}{R}$
$T_{max} = \frac{1}{R} \left( P_0 \sqrt{\frac{P_0}{3\alpha}} - \alpha \frac{P_0}{3\alpha} \sqrt{\frac{P_0}{3\alpha}} \right) = \frac{1}{R} \left( P_0 - \frac{P_0}{3} \right) \sqrt{\frac{P_0}{3\alpha}}$
$T_{max} = \frac{2P_0}{3R} \sqrt{\frac{P_0}{3\alpha}}$.

Solution:
Efficiency ($\eta$) = $\frac{\text{Work Output}}{\text{Heat Input}} = \frac{W}{Q_1} $
$\eta = \frac{1 \text{ kJ}}{5 \text{ kJ}} = 0.2$ or $20\%$.
By energy balance, $Q_2 = Q_1 - W = 5 - 1 = 4 \text{ kJ}$.
Heat rejected = 4 kJ.

Solution:
$ \alpha = \frac{Q_2}{W} \implies 5 = \frac{50}{W} \implies W = 10 \text{ J} $
Work required by compressor = 10 J.
Heat rejected to room ($Q_1$) = $Q_2 + W = 50 + 10 = 60 \text{ J}$.

Solution:
CRITICAL STEP: Convert to Kelvin.
$T_1 = 227 + 273 = 500 \text{ K}$
$T_2 = 27 + 273 = 300 \text{ K}$
Carnot Efficiency $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$.
Efficiency is 40%.

Solution:
Let initially $\eta_1 = 1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_1 = \frac{6}{5} T_2$. (Equation 1)
New efficiency is double, so $\eta_2 = \frac{2}{6} = \frac{1}{3}$. New sink temp is $(T_2 - 62)$.
$\eta_2 = 1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3} \implies T_1 = \frac{3}{2}(T_2 - 62)$. (Equation 2)
Equate $T_1$ from both equations:
$\frac{6}{5} T_2 = \frac{3}{2}(T_2 - 62)$
$12 T_2 = 15(T_2 - 62) \implies 12 T_2 = 15 T_2 - 930$
$3 T_2 = 930 \implies T_2 = 310 \text{ K}$.
From Eq 1: $T_1 = \frac{6}{5}(310) = 372 \text{ K}$.
Source $T_1 = 372 \text{ K}$, Sink $T_2 = 310 \text{ K}$.

Solution:
Case 1: Equal Work Outputs ($W_A = W_B$)
$W_A = Q_1 - Q_T$ and $W_B = Q_T - Q_2$.
Since $W_A = W_B \implies Q_1 - Q_T = Q_T - Q_2 \implies Q_T = \frac{Q_1 + Q_2}{2}$.
For Carnot engines, heat transfers are rigorously proportional to absolute temperatures: $\frac{Q_1}{T_1} = \frac{Q_T}{T} = \frac{Q_2}{T_2}$.
So substituting relations gives: $T = \frac{T_1 + T_2}{2}$.
$T = \frac{600 + 300}{2} = 450 \text{ K}$. (Arithmetic Mean)

Case 2: Equal Efficiencies ($\eta_A = \eta_B$)
$\eta_A = 1 - \frac{T}{T_1}$ and $\eta_B = 1 - \frac{T_2}{T}$.
$1 - \frac{T}{T_1} = 1 - \frac{T_2}{T} \implies \frac{T}{T_1} = \frac{T_2}{T} \implies T^2 = T_1 T_2$.
$T = \sqrt{T_1 T_2} = \sqrt{600 \times 300} = \sqrt{180000} \approx 424.26 \text{ K}$. (Geometric Mean)