Question

One mole of an ideal gas is heated at a constant pressure of $1atm$ 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
$\Delta Q=n{C}_P\Delta T$
Change in internal energy,
$\Delta U=n{C}_V\Delta T$
According to first law of thermodynamics
$\Delta Q=\Delta U+\Delta W$
$\therefore \Delta W=\Delta Q-\Delta U$
$=n{C}_P\Delta T-n{C}_v\Delta T$
$=n\Delta t\left(C_P-C_V\right)$
For an ideal gas, $C_P-C_V=R$
$\therefore \Delta W=nR\Delta T$
Here, $n=1$, $R=8.31Jmo{l}^{-1}{K}^{-1}$
$\Delta T=T_2-T_1=100^{\circ }C-0^{\circ }C=100^{\circ }C=100K$
$\therefore \Delta W=1\times 8.31\times 100$
$=8.31\times 10^{2}J$