The molar internal enegry of a gas over a temperature range is expressed as: `U_(m) (T) =a +bT +cT^(2)`, where `b = 16 J mol^(-1)K^(-1)`and `c = 6 xx 10^(-3) J mol^(-1)K^(-2)`. Find:
a. `C_(V,m)at 300K`
b. The entropy change when `1 mol` of a gas is heated from `300K` to `600K` at constant volume.

Explanation

a. Using: `dU_(m) = C_(V.m)dT`
`rArr C_(V,m) = (dU_(m))/(dT) = b +2cT`
`rArr C_(V.m) = 16 +2 xx6 xx 10^(-3) xx 300 = 19.6J mol^(-1)K^(-1)`
b. `dS = (dq)/(T) = (dU+(-w))/(T) = (dU+P dV)/(T) =(dU)/(T) [ :' dV = 0]`
`= (nC_(V)dT)/(T) = n(b +2cT). (dT)/(T)`
`rArr DeltaS = overset(T_(2))underset(T_(1))int (nbdT)/(T) + overset(T_(2))underset(T_(1))int 2nc dT`
`= nb "In" (T_(2))/(T_(1))+ 2nc (T_(2) - T_(1))`
`rArr DeltaS = 1 xx 16 xx "In"(600)/(300) +2 xx1xx6 xx 10^(-3) xx (600 - 300)`
`= 14.8 J mol^(-1) K^(-1)`