Question

Two moles of a rigid diatomic gas does a work \(\frac{Q}{4}\) when the amount of heat supplied is \(Q\). In this process the total heat capacity of the gas is (\(R\) = gas constant)

• A. \(\frac{12R}{5}\)
• B. \(\frac{10R}{7}\)
• C. \(\frac{15R}{4}\)
• D. \(\frac{20R}{3}\)

Hint:

Use the first law to get the change in internal energy, then express heat supplied in terms of the temperature change.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
First law: \(Q=\Delta U+W\).
Heat capacity of the gas in this process is \(C=\dfrac{Q}{\Delta T}\).

Step 2: Find the internal energy change
\(W=\dfrac{Q}{4}\), so \(\Delta U=Q-\dfrac{Q}{4}=\dfrac{3Q}{4}\).

Step 3: Use the diatomic gas value
For two moles of a rigid diatomic gas, \(\Delta U=nC_V\Delta T=2\times\dfrac{5R}{2}\Delta T=5R\Delta T\).

Step 4: Calculation
\(Q=\dfrac{4}{3}\Delta U=\dfrac{20R}{3}\Delta T\), so \(C=\dfrac{20R}{3}\), option (D).

Final Answer:
The total heat capacity of the gas is 20R by 3. \[ \boxed{\text{(D) }\dfrac{20R}{3}} \]