Question

Two moles of an ideal gas heated at constant pressure of one atmosphere from ${27}^{o}Cto{127}^{o}C.$ If $C_{v,m}=(20+{10}^{-2}T)J{K}^{-1}mo{l}^{-1},$ then $q$ and $△U$ 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

$△U$ (internal energy) is given as:
$dU=n{C}_{v}dT$
integrating both sides we get
$△U=n{\int }_{300}^{400}(20+{10}^{-2}T)dT=2[20\times 100+\frac{{10}^{-2}}{2}({400}^2-{300}^2\left)\right]=4700J$

we know at constant pressure heat absorbed is given as:
$dq=n{C}_{p}dT$
and $C_p=R+C_v$
so, $dq=n(C_v+R)dT$

integrating both sides we get :
$q=n{\int }_{300}^{400}(R+20+{10}^{-2}T)dT=2\left[\right(R\times 100)+(20\times 100)+\frac{{10}^{-2}}{2}({400}^2-{300}^2\left)\right]=6362.8J$