Question

If \(\Delta Q\) is the amount of heat supplied to 'n' moles of a rigid diatomic gas at constant pressure, \(\Delta U\) is the change in internal energy and \(\Delta W\) is the work done then \(\Delta W:\Delta U:\Delta Q\) is

• A. \(2:5:7\)
• B. \(1:4:7\)
• C. \(2:3:4\)
• D. \(3:5:9\)

Hint:

At constant pressure, Q = n Cp dT, dU = n Cv dT and W = n R dT.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a rigid diatomic gas, \(C_v=\dfrac52R\) and \(C_p=\dfrac72R\). At constant pressure the first law reads \(\Delta Q=\Delta U+\Delta W\).

Step 2: Write each quantity per unit n delta T:
\(\Delta Q=nC_p\Delta T=\dfrac72nR\Delta T\). \(\Delta U=nC_v\Delta T=\dfrac52nR\Delta T\). \(\Delta W=P\Delta V=nR\Delta T\).

Step 3: Take the ratio:
\(\Delta W:\Delta U:\Delta Q=1:\dfrac52:\dfrac72=2:5:7\). Option A.

Step 4: Why the other options are wrong.
The sum of the first two terms must equal the third. Only \(2+5=7\) holds. For example \(1+4\neq7\) in option B.

Final Answer:
The ratio is 2 : 5 : 7. \[ \boxed{\text{(A) }2:5:7} \]