As ideal monoatomic gas of 1.5 moles is heated at a constant pressure $2 \mathrm{~atm}$ so that its…
As ideal monoatomic gas of 1.5 moles is heated at a constant pressure $2 \mathrm{~atm}$ so that its temperature increases from $30^{\circ} \mathrm{C}$ to $130^{\circ} \mathrm{C}$ work done by the gas is (Universal gas constant $\left.=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
$2500 \mathrm{~J}$
$1450 \mathrm{~J}$
$1245 \mathrm{~J}$
$555 \mathrm{~J}$
Solution
Work done by a gas in a constant pressure process is
$W=n R \Delta T$...(i)
Here, $n=1.5$ moles
$\begin{aligned} R & =8.3 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1} \\ \Delta T & =\{(130+273)-(30+273)\}=100 \mathrm{~K}\end{aligned}$
So,from eq. (i)
$W=1.5 \times 8.3 \times 100=1245 J$