JEE Challenger
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PhyMedAdv 2021 • Paper 2

Ratio of Right Chamber Temperature to Initial Temperature

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 value of the ratio TRT0\frac{T_R}{T_0}.

Question Diagram 1

Options

A

2\sqrt{2}

Correct
B

3\sqrt{3}

C

22

D

33

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsKinetic Theory of Gases

Step-by-Step Solution

To determine the value of the ratio TRT0\frac{T_R}{T_0}, we analyze the thermodynamic process occurring in the right chamber of the cylinder.

Step 1: Determine the adiabatic index (γ\gamma) of the gas The molar heat capacity at constant volume is given as: CV=2RC_V = 2R

Using Mayer's relation (CP−CV=RC_P - C_V = R), the molar heat capacity at constant pressure is: CP=CV+R=2R+R=3RC_P = C_V + R = 2R + R = 3R

The adiabatic index (γ\gamma) for the gas is: γ=CPCV=3R2R=32\gamma = \frac{C_P}{C_V} = \frac{3R}{2R} = \frac{3}{2}

Step 2: Identify the process for the gas in the right chamber

  • The cylinder walls and the movable partition are thermally insulating, so no heat is transferred into or out of the right chamber (QR=0Q_R = 0).
  • The partition moves slowly (quasistatically) without friction.

Hence, the gas in the right chamber undergoes a reversible adiabatic compression.

Step 3: Apply the adiabatic relation for temperature and volume For a reversible adiabatic process, the relationship between temperature TT and volume VV is given by: TVγ−1=constantT V^{\gamma - 1} = \text{constant}

Substituting γ=32\gamma = \frac{3}{2}, we get: γ−1=32−1=12\gamma - 1 = \frac{3}{2} - 1 = \frac{1}{2}

So, the equation becomes: TV=constantT \sqrt{V} = \text{constant}

Step 4: Calculate the final temperature ratio TRT0\frac{T_R}{T_0}

  • Initial state of the right gas: Temperature = T0T_0, Volume = V0V_0
  • Final state of the right gas: Temperature = TRT_R, Volume = V02\frac{V_0}{2}

Applying the relation between the initial and final states: T0V0=TRV02T_0 \sqrt{V_0} = T_R \sqrt{\frac{V_0}{2}}

Rearranging to solve for TRT0\frac{T_R}{T_0}: TRT0=V0V02=2\frac{T_R}{T_0} = \frac{\sqrt{V_0}}{\sqrt{\frac{V_0}{2}}} = \sqrt{2}

Thus, the value of the ratio TRT0\frac{T_R}{T_0} is 2\sqrt{2}.

Correct Option: (A)

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