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: Use the first law as a check:
$\Delta Q=\Delta U+\Delta W$, so the third number must be the sum of the other two.

Step 2: Use the ratio of specific heats:
$\dfrac{\Delta U}{\Delta Q}=\dfrac{C_v}{C_p}=\dfrac1\gamma=\dfrac57$. So $\Delta W/\Delta Q=1-\dfrac57=\dfrac27$.

Step 3: Write the ratio:
$\Delta W:\Delta U:\Delta Q=2:5:7$. Option A.

Final Answer:
Using Cp = 7R/2 and Cv = 5R/2 gives W : U : Q = 2 : 5 : 7. \[ \boxed{\text{(A) }2:5:7} \]