Question

Two mole of an ideal gas is heated at constant pressure of one atomosphere 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 respectively.

• A. 6362.8 J, 4700 J
• B. 3037.2 J, 4700 J
• C. 7062.8, 5400 J
• D. 3181.4 J, 2350 J

Answer 1

The correct option is A 6362.8 J, 4700 J

$q=\int C_{V}dT=\int (20+{10}^{-2}T)\times 2dt=2[20T+\frac{{10}^{-2}{T}^2}{2}]=2[20(400-300)+\frac{{10}^{-2}}{2}[{400}^2-{300}^2]]=2\left[2350\right]=4700J△U=q+W$

Work done in isobaric process is $P△V$

$\left(1\right)\to 1\times V_1=2\times 8.314\times 300[PV=nRT]\left(2\right)\to 1\times V_2=2\times 8.314\times 400V_2-V_1=2\times 8.314\times 100=1662.8W=P△V=1\times 1662.8=1662.8J△U=q+W=4700+1662.8=6362.8J$