The constant-pressure heat capacity of a sample of a perfect gas was f

Step 1
(b) In this excercise we have to calculate q, w, $\mathrm{△}Uand\mathrm{△}H$ when the temperature is raised from ${25}^{\circ }Cto{100}^{\circ }C$ at constant pressure and at constant volume.
(i)
At constant pressure:
First step is to convert ${}^{\circ }CtoK.$
We can do it when we add value of degree centigrade to 273.15, so:
$T(K)=t{(}^{\circ }C)+273.15$
For ${25}^{\circ }C:$
$T_i(K)=25+273.15$
$=298.15$
For ${100}^{\circ }C:$
$T_f(K)=100+273.15$
$=373.15$
In case of constant pressure $q_p=\mathrm{△}H$ so we can express $\mathrm{△}H$ as:
$$\begin{gathered} \mathrm{△}H={\int }_{T_i}^{T_f}dH \\ ={\int }_{T_i}^{{\tau }_T}{C}_{p}dT \\ ={\int }_{273.15K}^{373.15K}\{20.17J{K}^{-1}+0.4001J{K}^{-1}T\}dT \\ =20.17J{K}^{-1}[T{]}_{273.15}^{373.15}+\frac{0.4001}{2}[T^2{]}_{273.15}^{373.15} \\ =20.17J{K}^{-1}[373.15K-273.15K]+\frac{0.4001J{K}^{-2}}{2}[(373.15K{)}^2-(273.15K{)}^2] \\ =1.16\cdot {10}^{4}J \end{gathered}$$
Step 2
Now we have to convert change in enthalpy to kilojoules. if we know that 1kJ=1000J
$\mathrm{△}H=1.16\cdot {10}^{4}J\cdot \frac{1kJ}{1000J}$
$=+11.6kJ$
Finally, $\mathrm{△}H=+11.6kJ$
As $q_p=\mathrm{△}H$ at constant pressure, we now also know value for q and it is equal so:
$q_p=+11.6kJ$
Next step is work done:
NSK
At constant pressure work done can be expressed as:
$\omega =-p\mathrm{△}V$
In this case $p=p_{ex}$ because external pressure is constant
$$\begin{gathered} \omega =-p_{ex}\mathrm{△}V \\ =-\mathrm{△}pV \\ =-\mathrm{△}(nRT) \\ =-nR\mathrm{△}T \\ =-(1.0mol)(8.314J{K}^{-1}mo{l}^{-1})(373.15K-298.15K) \\ =-623.55J \\ =-0.624kJ \end{gathered}$$
So work done is equal to -0.624kJ
Step 3
(ii)
At constant volume
In this case change in internal energy and enthalpy of a perfect gas is equal to these two at constant pressure because they depends just on temperature
NSK
Therefore:
$\mathrm{△}H=+11.6kJ$
Change in internal energy:
$$\begin{gathered} \mathrm{△}U=q+\omega \\ \omega =0-\text{work done at constant volume so}\mathrm{△}U=q \\ \mathrm{△}U=q+\omega \\ =(11.6kJ)+(-0.624kJ) \\ =+11.0kJ \end{gathered}$$
As $\omega =0$, we come to a conclusion that $q=+11.0kJ$