Two cylinders \(\mathrm{A}\) and \(\mathrm{B}\) fitted with piston contain equal amounts of an ideal diatomic gas at \(300 \mathrm{k}\). The piston of \(\mathrm{A}\) is free to move, While that of \(B\) is held fixed. The same amount of heat is given to the gas in each cylinders. If the rise in temperature of the gas in \(\mathrm{A}\) is \(30 \mathrm{~K}\), then the rise in temperature of the gas in \(\mathrm{B}\) is. (A) \(30 \mathrm{~K}\) (B) \(42 \mathrm{~K}\) (C) \(18 \mathrm{~K}\) (D) \(50 \mathrm{~K}\)

To find the change in internal energy, we use the formula: \[ \Delta U = n C_v \Delta T \] where \(\Delta U\) is the change in internal energy, \(n\) is the number of moles, \(C_v\) is the molar specific heat at constant volume, and \(\Delta T\) is the change in temperature. For an ideal diatomic gas, we have the following values: - \(C_v = \frac{5}{2} R\), where \(R\) is the universal gas constant Now, the change in temperature for cylinder A is given as \(\Delta T_A = 30\) K. The change in internal energy or \(\Delta U_A\) will be: \[ \Delta U_A = n C_v \Delta T_A = n\left(\frac{5}{2} R\right)(30 K) \]

Using the first law of thermodynamics, we have: \[ q_A = \Delta U_A + W_A \] where \(q_A\) is the heat given to cylinder A, \(\Delta U_A\) is the change in internal energy, and \(W_A\) is the work done by the gas in cylinder A. We know the heat given to both cylinders is equal. So, we have: \[ W_A = q_A - \Delta U_A \]

Since the piston in cylinder B is held fixed, no work is done, and we have: \[ \Delta U_B = q_B \] As the heat given to both cylinders is equal, we get: \[ \Delta U_B = q_B = q_A \]

We know that \(\Delta U_B = n C_v \Delta T_B\). We can use this to find the change in temperature for cylinder B, denoted as \(\Delta T_B\): \[ \Delta T_B = \frac{\Delta U_B}{nC_v} = \frac{q_A}{n\left(\frac{5}{2}R\right)} \] Since \(q_A = \Delta U_A + W_A\), we can substitute this in the formula for \(\Delta T_B\): \[ \Delta T_B = \frac{\Delta U_A + W_A}{n\left(\frac{5}{2}R\right)} = \frac{n\left(\frac{5}{2}R\right)(30 K) + W_A}{n\left(\frac{5}{2} R\right)} \] Simplifying, we get: \[ \Delta T_B = 30 K + \frac{W_A}{n\left(\frac{5}{2}R\right)} \]

Now we need to determine the exact value of \(\Delta T_B\). First, we need to find the value of \(W_A\). From step 2, we have: \[ W_A = q_A - \Delta U_A = q_A - n\left(\frac{5}{2} R\right)(30 K) \] Since \(q_A = \Delta U_B\), we have: \[ W_A = \Delta U_B - n\left(\frac{5}{2} R\right)(30 K) = n\left(\frac{5}{2} R\right)(\Delta T_B - 30 K) \] Plugging this into the formula for \(\Delta T_B\): \[ \Delta T_B = 30 K + \frac{n\left(\frac{5}{2} R\right)(\Delta T_B - 30 K)}{n\left(\frac{5}{2} R\right)} \] Which simplifies to: \[ \Delta T_B = 30 K + (\Delta T_B - 30 K) \] This gives us: \[ \Delta T_B = 30 K \] So the rise in temperature of the gas in cylinder B is the same as in cylinder A, which is 30 K (Option A).