Looking for classes? Ksquare Career Institute, Bengaluru →
2 mole of ideal gas at $27^\circ C$ temp. is expanded reversibly from 2 lit. to 20 lit. Find entropy change (R = 2 cal/mol K) : -
A
92.1
B
0
C
4
D
9.2
Detailed Solution
For reversible isothermal expansion of an ideal gas: $\Delta S = nR\ln\dfrac{V_2}{V_1} = 2.303\,nR\log\dfrac{V_2}{V_1}$
$\Delta S = 2.303 \times 2 \times 2 \times \log\dfrac{20}{2}$
$= 2.303 \times 4 \times \log 10 = 2.303 \times 4 \times 1$
$\Delta S = 9.2$ cal K$^{-1}$
(The temperature does not enter the calculation, since the process is isothermal.)
$\Delta S = 2.303 \times 2 \times 2 \times \log\dfrac{20}{2}$
$= 2.303 \times 4 \times \log 10 = 2.303 \times 4 \times 1$
$\Delta S = 9.2$ cal K$^{-1}$
(The temperature does not enter the calculation, since the process is isothermal.)