Question

When $1$ mol of an ideal gas is compressed to half of its volume, its temperature becomes double; then the change in entropy ($\Delta S$) would be:

• A. $C_V\ln 2$
• B. $C_P\ln 2$
• C. $C_{V}R\ln 2$
• D. $(C_V-R)\ln 2$

Answer 1

Answer: (D)

$(C_V-R)\ln 2$

The expression of entropy is given as

$\Delta S=2.303C_{v}log\frac{T_2}{T_1}+2.303Rlog\frac{V_2}{V_1}$

we have $T_2=2T_1$ and $V_1=2V_2$

so we get,

$\Delta S=2.303C_{v}log\frac{2}{1}+2.303Rlog\frac{1}{2}$

or we can write it as

$\Delta S=(C_v-R)ln2$