Question

Two moles of an ideal gas is heated at constant pressure of one atomsphere from ${27}^{\circ }C$ to ${127}^{\circ }C.$ if $C_{v,m}=20+{10}^{-2}T$ $J{K}^{-1}mo{l}^{-1}$ then Heat (q) and Internal energy ($△U$) for the process are respectively:

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

Answer 1

The correct option is A 6362.8 J, 4700 J

Calculating $w$ using :
$w=-P△V=-nR△T$ (for an ideal gas)
$w=-nR△T=-2\times 8.314\times 100=-1662.8J$

now molar heat capacity at constant volume is given as
$C_v=\frac{dU}{ndT}$ or

$dU=n{C}_{v}dT$
given, $C_v=20+{10}^{-2}T$
$T_1=300K$ and $T_2=400K$ so,
integrating both sides from $T_1$ to $T_2$ :

$△U=n{\int }_{300}^{400}(20+{10}^{-2}T)dT=2[20\times T+\frac{{10}^{-2}}{2}(T^2\left)\right]=2[20\times (T_2-T_1)+\frac{{10}^{-2}}{2}(T_2^2-T_1^2\left)\right]=2[20\times 100+\frac{{10}^{-2}}{2}({400}^2-{300}^2\left)\right]So,△U=4700J$

also $△U=q+w$
$4700=q-1662.8\therefore q=6362.8J$