JEE Challenger
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PhyHardAdv 2021 • Paper 2

Dimensionless Heat Supplied to Gas Chamber

Comprehension Passage

A thermally insulated cylinder is divided into two parts by a frictionless and thermally insulating movable partition. Each section initially contains one mole of an ideal gas with constant volume molar heat capacity CV=2RC_V = 2R. Here, RR represents the gas constant. Initially, both sides have volume V0V_0 and temperature T0T_0. An electric heater on the left side is activated at very low power, supplying heat QQ to the left gas. Consequently, the partition moves slowly to the right, reducing the volume of the right chamber to V0/2V_0/2. The final temperatures of the gases on the left and right sides are TLT_L and TRT_R, respectively. Temperature changes of the cylinder, partition, and heater may be neglected.

Determine the dimensionless value of QRT0\frac{Q}{RT_0}.

Question Diagram 1

Options

A

4(22+1)4(2\sqrt{2} + 1)

B

4(22−1)4(2\sqrt{2} - 1)

Correct
C

(52+1)(5\sqrt{2} + 1)

D

(52−1)(5\sqrt{2} - 1)

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsHeat Transfer

Step-by-Step Solution

To determine the dimensionless value of QRT0\frac{Q}{RT_0}, we apply the principles of thermodynamics to both the left and right gas chambers.

1. Initial Parameters

Each chamber contains n=1 molen = 1\text{ mole} of ideal gas with constant volume molar heat capacity: CV=2RC_V = 2R

The adiabatic exponent γ\gamma for this gas is: γ=CPCV=CV+RCV=2R+R2R=32\gamma = \frac{C_P}{C_V} = \frac{C_V + R}{C_V} = \frac{2R + R}{2R} = \frac{3}{2}

Initially, both chambers have:

  • Volume: VL=VR=V0V_L = V_R = V_0
  • Temperature: TL=TR=T0T_L = T_R = T_0
  • Pressure: P0=RT0V0P_0 = \frac{RT_0}{V_0}

2. Process for the Right Chamber

Since the cylinder walls and the partition are thermally insulating and the process occurs very slowly (quasistatically), the gas in the right chamber undergoes a reversible adiabatic compression.

The final volume of the right chamber is VR′=V02V_R' = \frac{V_0}{2}.

Using the adiabatic relation TVγ−1=constantTV^{\gamma - 1} = \text{constant}: T0V0γ−1=TR(VR′)γ−1T_0 V_0^{\gamma - 1} = T_R (V_R')^{\gamma - 1} TR=T0(V0V0/2)32−1=T0(2)1/2=2T0T_R = T_0 \left(\frac{V_0}{V_0/2}\right)^{\frac{3}{2} - 1} = T_0 (2)^{1/2} = \sqrt{2}T_0

Using the adiabatic relation PVγ=constantPV^\gamma = \text{constant} to find the final equilibrium pressure PfP_f: P0V0γ=Pf(VR′)γP_0 V_0^\gamma = P_f (V_R')^\gamma Pf=P0(V0V0/2)3/2=P0(2)3/2=22P0P_f = P_0 \left(\frac{V_0}{V_0/2}\right)^{3/2} = P_0 (2)^{3/2} = 2\sqrt{2} P_0

3. Process for the Left Chamber

Since the total volume of the insulated cylinder remains 2V02V_0, the final volume of the left chamber is: VL′=2V0−VR′=2V0−V02=3V02V_L' = 2V_0 - V_R' = 2V_0 - \frac{V_0}{2} = \frac{3V_0}{2}

Since the partition is frictionless, the final pressure on the left side must equal the final pressure on the right side: PL′=Pf=22P0P_L' = P_f = 2\sqrt{2} P_0

Applying the ideal gas equation for the left chamber in its final state: PL′VL′=nRTLP_L' V_L' = n R T_L (22P0)(3V02)=1⋅RTL(2\sqrt{2} P_0) \left(\frac{3V_0}{2}\right) = 1 \cdot R T_L TL=32P0V0R=32T0T_L = \frac{3\sqrt{2} P_0 V_0}{R} = 3\sqrt{2} T_0

4. Heat Supplied (QQ) via First Law of Thermodynamics

Considering both gas chambers together as a single composite system:

  • The external work done by the system is zero (Wext=0W_{\text{ext}} = 0) because the outer boundaries are rigid.
  • Heat QQ is supplied only to the left chamber from the electric heater.

By the First Law of Thermodynamics: Q=ΔUtotal=ΔUL+ΔURQ = \Delta U_{\text{total}} = \Delta U_L + \Delta U_R

Calculating the change in internal energy for each side: ΔUL=nCV(TL−T0)=1⋅(2R)(32T0−T0)=2RT0(32−1)\Delta U_L = n C_V (T_L - T_0) = 1 \cdot (2R)(3\sqrt{2}T_0 - T_0) = 2RT_0(3\sqrt{2} - 1) ΔUR=nCV(TR−T0)=1⋅(2R)(2T0−T0)=2RT0(2−1)\Delta U_R = n C_V (T_R - T_0) = 1 \cdot (2R)(\sqrt{2}T_0 - T_0) = 2RT_0(\sqrt{2} - 1)

Summing these changes: Q=2RT0(32−1)+2RT0(2−1)Q = 2RT_0(3\sqrt{2} - 1) + 2RT_0(\sqrt{2} - 1) Q=2RT0[(32−1)+(2−1)]Q = 2RT_0 \left[(3\sqrt{2} - 1) + (\sqrt{2} - 1)\right] Q=2RT0(42−2)=4RT0(22−1)Q = 2RT_0 (4\sqrt{2} - 2) = 4RT_0 (2\sqrt{2} - 1)

Rearranging to get the dimensionless ratio: QRT0=4(22−1)\frac{Q}{RT_0} = 4(2\sqrt{2} - 1)

Correct Answer:

B

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