Question

Half mole of an ideal monoatomic gas is heated at constant pressure of $1\text{atm}$ from ${20}^{\circ }\text{C}\text{to}{90}^{\circ }\text{C}$. Work done by gas is close to: (Gas constant $R=8.31\text{J/mol-K}$)

• A. $581\text{J}$
• B. $291\text{J}$
• C. $146\text{J}$
• D. $73\text{J}$

Answer 1

The correct option is B $291\text{J}$

Work done for a isobaric process is
$W=P\Delta V=nR\Delta T=\frac{1}{2}\times 8.31\times 70\approx 291\text{J}$.