Question

Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is

• A. 30 K
• B. 18 K
• C. 50 K
• D. 42 K

Answer 1

The Correct Option is D

A is free to move, therefore, heat will be supplied at constant
pressure
$\therefore \, \, \, \, \, \, \, dQ_A=nC_p dt_A \, \, \, \, \, \, \, \, \, \, \, \, $ ...(i)
B is held fixed, therefore, heat will be supplied at constant
volume.
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, dQ_B=nC_vdT_B \, \, \, \, \, \, \, \, \, \, \, \, $ ...(ii)
But $ \, \, \, \, \, \, \, \, \, \, \, \, dQ_A =dQ_B \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ (given)
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, _nC_pdt_A=nCvdT_B$
$\therefore \, \, \, \, \, \, \, \, \, \, dT_B=\big(\frac{C_p}{C_v}\bigg)dT_A$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\lambda(dT_A) \, \, \, \, \, [\gamma =1.4 (diatomic)]$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (dT_A=30 K)$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =(1.4)(30 K)$
$\therefore \, \, \, \, \, \, \, \, \, \, dT_B=42 K$

Answer 2

To solve this problem, we need to consider the differences in the behavior of the gases in the two cylinders A and B due to the constraints on their pistons. Let's denote the initial amount of heat given to both gases as \(Q\).

Cylinder A: Isobaric Process (Piston Free to Move)
Since the piston in cylinder A is free to move, the process is isobaric (constant pressure). For an ideal diatomic gas, the molar specific heat capacity at constant pressure \(C_p\) is given by:
\[C_p = \frac{7}{2}R\]
where \(R\) is the universal gas constant.

The heat added to the gas in cylinder A causes a temperature rise of \(30 \text{ K}\). We can use the formula for heat added at constant pressure:
\[Q = n C_p \Delta T_A\]
where \(n\) is the number of moles of the gas and \(\Delta T_A\) is the temperature rise in cylinder A.

Given \(\Delta T_A = 30 \text{ K}\), we have:
\[Q = n \left(\frac{7}{2}R\right) \times 30\]

Cylinder B: Isochoric Process (Piston Fixed)
In cylinder B, the piston is held fixed, so the process is isochoric (constant volume). For an ideal diatomic gas, the molar specific heat capacity at constant volume \(C_v\) is given by:
\[C_v = \frac{5}{2}R\]

The same amount of heat \(Q\) is given to the gas in cylinder B, leading to a temperature rise of \(\Delta T_B\). We can use the formula for heat added at constant volume:
\[Q = n C_v \Delta T_B\]

Substituting the expression for \(Q\) from the isobaric process into the isochoric process:
\[n \left(\frac{7}{2}R\right) \times 30 = n \left(\frac{5}{2}R\right) \Delta T_B\]

Solving for \(\Delta T_B\):
\[\left(\frac{7}{2}R\right) \times 30 = \left(\frac{5}{2}R\right) \Delta T_B\]

Dividing both sides by \(\left(\frac{5}{2}R\right)\):
\[30 \times \frac{7}{5} = \Delta T_B\]

\[\Delta T_B = 42 \text{ K}\]

Therefore, the rise in temperature of the gas in cylinder B is option (D) \(42 \text{ K}\).