Question

Consider an isothermal reversible compression of 1 mole of an ideal gas in which pressure of the system increased from 5 atm to 30 atm at 300 K. The entropy change (in kJ) of the surrounding is:
($R=8.314Jmo{l}^{-1}{K}^{-1}$)

• A. $1.5kJ/mol$
• B. $15kJ/mol$
• C. $0.07kJ/mol$
• D. $0.7kJ/mol$

Answer 1

The correct option is B $15kJ/mol$

We know, $dW=-pdV$
Then
$W=-{\int }_{V_1}^{V_2}PdV=-{\int }_{V_1}^{V_2}\frac{nRT}{V}dV=-nRTln\left(\frac{V_1}{V_2}\right)$

or, $W=-nRTln\left(\frac{P_1}{P_2}\right)$ [$\because \frac{P_1}{P_2}=\frac{V_1}{V_2}$]

or, $W=-1\times 8.314\times 300\times ln\left(\frac{5}{30}\right)=-8.314\times 300\times 1.79$

or, $W=-4.46kJ$
Therefore, $Q=-W=4.46kJ$ [$\because \Delta U=0$]

$\Delta S=\frac{Q}{T}=\frac{4.46\times {10}^3}{3\times {10}^2}=15kJ/mol$