Question

One mole of an ideal gas is heated at constant pressure of 1 atm from $0^o$C to ${100}^o$C . Work done by the gas is

• A. 8.31 $\times {10}^{3}J$
• B. 8.31 $\times {10}^{-3}J$
• C. 8.31 $\times {10}^{-2}J$
• D. 8.31 $\times {10}^{2}J$

Answer 1

The correct option is D 8.31 $\times {10}^{2}J$

Given,

$n=1$

$T_1=0^{0}C=273K$

$T_2={100}^{0}C=373K$

$P=1atm$

From ideal gas equation,

$PV=nRT$

Differentiate both side,

$P\Delta V=nR\Delta T$

Work done, $W=-P\Delta V=-nR(T_1-T_2)$

$W=-1\times 8.31\times (273-373)$

$W=831J=8.31\times {10}^{2}J$

The correct option is D.