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

Answer 1

The correct option is B $3818$ J

Work done in isothermal reaction is-

$W=-nRTln\frac{P_1}{P_2}$

$W=-.1\times 8.314\times 1000\times ln\left[{10}^{-2}\right]$

$W=-3818J$