Question

A rigid diatomic gas undergoes adiabatic change. Its pressure P and temperature T are related as \(P\propto T^x\) where x is

• A. \(1.5\)
• B. \(2.5\)
• C. \(3.5\)
• D. \(4.5\)

Hint:

Use T^gamma P^(1-gamma) = constant with gamma = 7/5 for a rigid diatomic gas.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For an adiabatic process of an ideal gas, \(PV^\gamma=\) constant and \(PV=nRT\). Eliminating \(V\) gives \(P^{1-\gamma}T^{\gamma}=\) constant.

Step 2: Find x:
\(P\propto T^{\gamma/(\gamma-1)}\), so \(x=\dfrac{\gamma}{\gamma-1}\).

Step 3: Find gamma for a rigid diatomic gas:
\(C_v=\dfrac52R\), \(C_p=\dfrac72R\), so \(\gamma=\dfrac75\). Then \(x=\dfrac{7/5}{2/5}=\dfrac72=3.5\). Option C.

Step 4: Why the other options are wrong.
For a monatomic gas with \(\gamma=\tfrac53\), the exponent is \(\dfrac{5/3}{2/3}=2.5\), which is option B. Values 1.5 and 4.5 do not match any ideal gas.

Final Answer:
x = 3.5. \[ \boxed{\text{(C) }3.5} \]