Question

Calculate $\Delta U$ and $\Delta H$ in calories respectively, if one mole of a monoatomic ideal gas is heated at constant pressure of 1 atm from ${25}^{o}C$ to ${50}^{o}C$.

• A. $74.5caland124.2cal$
• B. $74.5caland49.5cal$
• C. $49.7caland-74.5cal$
• D. $-49.7caland-74.5cal$

Answer 1

The correct option is A $74.5caland124.2cal$

Given : $T_1={25}^{o}C=25+273=298K$ $T_2={50}^{o}C=50+273=323K$, $p=1atm$
For monoatmic ideal gas, $C_v=\frac{3}{2}R,C_p=\frac{5}{2}R$
But $C_p=\frac{\Delta H}{\Delta T}$
$\therefore \Delta H=n{C}_p\Delta T$ for n moles
$\Delta H=\frac{5}{2}\times 1.987\times 25=124.2cal$

work done, $w=-p\Delta V=-p(V_2-V_1)=-(p{V}_2-p{V}_1)=-(nR{T}_2-nR{T}_1)=-nR(T_2-T_1)$
$=-1\times 1.987(323-298)$
$=-49.7cal$
$\Delta U=q+w$
$\Delta U=124.2-49.7$
$\Delta U=74.5cal$