Question

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

• A. 6362.80 J, 4700 J
• B. 3037.20 J, 4700 J
• C. 7062.80 J, 5400 J
• D. 3181.40 J, 2350 J

Answer 1

The correct option is A 6362.80 J, 4700 J

We know, $-P\Delta V=W$ (at constant pressure)
So, $W=-n_{g}R\Delta T$
$\Rightarrow W=-2\times 8.314\times 100=-1662.80J$
Now internal energy, $\Delta U=n{\int }_{T_1}^{T_2}{C}_{v}dt$
$\Delta U=n{\int }_{300}^{400}(20+{10}^{-2}T)dT$
$\Rightarrow \Delta U=2[20\times 100+\frac{{10}^{-2}}{2}({400}^2-{300}^2\left)\right]=4700J$
From the first law of thermodynamics,
$4700=q-1662.80$
$\Rightarrow q=4700+1662.80=6362.80J$