Question

20 g helium is compressed isothermally and reversibly at 27 °C from a pressure of 2 atmosphere to 20 atmosphere. The change in the value of heat energy change during the process is (R = 2 cal kelvin-1 mol-1)

• A. (A) 83.14 kcal
• B. (B) -6.9 kcal
• C. (C) -20.7 kcal
• D. (D) 6.9 kcal

Answer 1

Answer: (D)

(D) 6.9 kcal

Answer 2

Explanation:
dU=dq+dW………(i)as process is isothermal dT=0so dU=nRdT=0so dq=−dWnow PV=nRTfor isothermal process PV=constant⇒PdV+VdP=0⇒PdV=−VdP………………..now dW=PdVso,dq=−PdVchange in heat energy Δq=∫dq=∫P1P2−PdV⇒Δq=∫P1P2VdPΔq=∫P1P2nRTPdPΔq=nRT∫P1P2dPP=nRTln⁡P2P1=−wMRTlne⁡P1P2(n=weightMolecular weight)Δq=−2.303wMRTlog10⁡P1P2Δq=−2.303204×2×300×log10⁡220=6909cal=6.9k.cal