Question

The minimum work which must be done to compress $0.1$ moles of an ideal gas, isothermally (at ${727}^{o}C$), from $1.04\times {10}^{4}N/m^2$ to $1.04\times {10}^{6}N/m^2$, is: $[\ln 10=2.3,R=8.3J/K-mol]$

• A. $3181J$
• B. $3818$
• C. $821.7J$
• D. $830J$

Answer 1

The correct option is A $3181J$

During compression reversible work done is the minimum work done.

$W_{rev}=nRTIn\frac{P_2}{P_1}$

$=nRTIn\frac{1.04\times {10}^6}{1.04\times {10}^{14}}$

$=0.1\times 8.3\times 1000m100$

$=3181J$