Question

2.0 moles of an ideal monoatomic gas is expanded isothermally from $10L$ to $20L$, at $27{}^{\circ }C$, against a constant external pressure of $0.1bar$. The correct changes in state parameters of gas in this process is/ are $[ln2=0.7,R=8.3J/K.mol]$

• A. $\Delta U_{sys}=0$
• B. $\Delta S_{sys}=11.62J/K$
• C. $\Delta S_{surr}=-0.33J/K$
• D. $\Delta G_{sys}=-3486J$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\Delta U_{sys}=0$
B $\Delta S_{sys}=11.62J/K$
C $\Delta S_{surr}=-0.33J/K$
D $\Delta G_{sys}=-3486J$

Since the process is isothermal, thus $\Delta U_{system}=0$

For an isothermal process,
$\Delta S_{system}=nRln\frac{V_2}{V_1}$
$=2\times 8.3\times ln2$
$=2\times 8.3\times 0.7$
$=11.62J/K$

Again,
$\Delta S_{surr}=\frac{q_{surr}}{T}=\frac{-q_{system}}{T}$
$=\frac{-(-W_{system})}{T}$
(Since the process is isothermal)
$\therefore \Delta S_{surr}=-\frac{P_{ext(V_2-V_1)}}{T}$
$=-\frac{0.1bar\times 10L}{300}$$=\frac{1\times 1000}{300}barml/K$
$=-\frac{1}{3}J/K=-0.33J/K$

We know,
$\Delta G=VdP-SdT(dT=0)$
$=\frac{nRT}{P}dP$
$=nRTIn\frac{P_2}{P_1}$

$=nRTIn\frac{V_1}{V_2}$
$=2\times 8.3\times 300In\frac{10}{20}$
$=-3486J/K$