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

Answer 1

The Correct Option is D

Explanation:
$dU=dq+dW\dots \dots \dots (i)$as process is isothermal $dT=0$so $dU=nRdT=0$so $dq=-dW$now $PV=nRT$for isothermal process $PV=$constant$\Rightarrow PdV+VdP=0$$\Rightarrow PdV=-VdP\dots \dots \dots \dots \dots \dots ..$now $dW=PdV$$so,dq=-PdV$change in heat energy $\mathrm{\Delta }q=\int dq={\int }_{P_1}^{P_2}-PdV$$\Rightarrow \mathrm{\Delta }q={\int }_{P_1}^{P_2}VdP$$\mathrm{\Delta }q={\int }_{P_1}^{P_2}\frac{nRT}{P}dP$$\mathrm{\Delta }q=nRT{\int }_{P_1}^{P_2}\frac{dP}{P}=nRT\ln \frac{P_2}{P_1}=-\frac{w}{M}RT{\ln }_e\frac{P_1}{P_2}(n=\frac{\text{weight}}{\text{Molecular weight}})$$\mathrm{\Delta }q=-2.303\frac{w}{M}RT{\log }_{10}\frac{P_1}{P_2}$$\mathrm{\Delta }q=-2.303\frac{20}{4}\times 2\times 300\times {\log }_{10}\frac{2}{20}$$=6909\mathrm{cal}=6.9\mathrm{k}.\mathrm{cal}$