Question

1 mole of ideal diatomic gas is enclosed in a cylinder piston arrangement having cross-sectional area of piston \(4\,\text{cm}^2\). If gas has only rotational modes and \(P_{atm}=100\ \text{kPa}\), some amount of heat is added to the system as a result piston moves up slowly by \(2.5\,\text{cm}\). If temperature change is \(1.2^\circ C\). Find heat given to gas.

• A. \(19.9\,J\)
• B. \(23.5\,J\)
• C. \(14.6\,J\)
• D. \(10\,J\)

Answer 1

The Correct Option is A

Concept: Since the piston moves slowly, the process is isobaric (pressure constant). Heat supplied: \[ Q = nC_p\Delta T \] For a diatomic gas with only rotational modes: \[ f = 2 \] \[ C_p = \left(\frac{f}{2}+1\right)R \]
Step 1:Find molar heat capacity at constant pressure. \[ C_p = \left(\frac{2}{2}+1\right)R \] \[ C_p = 2R \]
Step 2:Calculate heat supplied. \[ Q = nC_p\Delta T \] \[ Q = 1 \times 2R \times 1.2 \] \[ Q = 2 \times 8.314 \times 1.2 \] \[ Q = 19.95\,J \] \[ \boxed{Q \approx 19.9\,J} \]