Question

Two moles of an ideal gas is heated at a constant pressure of one atmosphere from ${27}^{\circ }C$ to ${127}^{\circ }C$. If $C_{v,m}=20+{10}^{-2}TJ{K}^{-1}mo{l}^{-1}$, then $q$ and $\Delta U$ for the process are:

• A. 6362.8 J, 4700 J
• B. 1522.2 J, 1124.4 J
• C. 7062.8, 5400 J
• D. 3181.4 J, 2350 J

Answer 1

The correct option is A

6362.8 J, 4700 J

First thing to realise here is that this is an isobaric process. Hence, $W=pdV=nRdT$
$W=-nR\Delta T=-2\times 8.314\times 100=-1662.85$

From the ideas learned in heat capacity, we know that $dU=mCdT$. On integration, you get:

$\Delta U=n\int C_{v,m}dT=2\int (20+{10}^{-2}T)dT$

= $2\times 20\times (T_2-T_1)+2\times {10}^{-2}\times \left(\frac{T_2^2-T_1^2}{2}\right)=4700J$

$\Delta U=W\Delta U=q-W$

$4700J=q-1662.8$

$q=6362.85$

Tip: You can use $R=25/3$ to make quicker calcuations.
Remember this works only if the units are $J/\left(molK\right)$