Question

One mole of an ideal gas is heated at a constant pressure of $1 \,atm $ from $0^{\circ} C$ to $100^{\circ}C$. 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

Answer: (D)

Heat supplied to the gas at constant pressure is
$\Delta Q=nC_{P}\Delta T$
Change in internal energy,
$\Delta U=nC_{V}\Delta T$
According to first law of thermodynamics
$\Delta Q=\Delta U+\Delta W$
$\therefore \Delta W=\Delta Q-\Delta U$
$=nC_{P}\Delta T-nC_{v}\Delta T$
$=n\Delta t\left(C_{P}-C_{V}\right)$
For an ideal gas, $C_{P}-C_{V}=R$
$\therefore \Delta W=nR\Delta T$
Here, $n=1$, $R=8.31\,J\,mol^{-1}\,K^{-1}$
$\Delta T=T_{2}-T_{1}=100^{\circ}C-0^{\circ}C=100^{\circ}C=100\,K$
$\therefore \Delta W=1\times 8.31\times 100$
$=8.31\times10^{2}\,J$