Question

A 1.50 mole sample of an ideal gas is allowed to expand adiabatically and reversibly to twice its original volume. In the expansion the temperature dropped from 296K to 239K . Calculate $\text{ΔE\hspace{0.05em}and\hspace{0.05em}ΔH}$ for the gas expansion.

Short answer

The value of change in internal energy is -230 KJ, and the change in enthalpy is -3.01 KJ.

Step-by-step solution

Definition of change in internal energy and change in enthalpy

The sum of the heat transferred and the work done is defined as a change in internal energy.

Approximately equal to the difference between the energy used to break bonds in a chemical reaction and the energy gained by the formation of new chemical bonds in the reaction is defined as the enthalpy change.

Calculations

Initial temperature${\text{(T}}_{\text{1}}\text{)}=296\text{K}$

Final temperature${\text{(T}}_2\text{)}=296\text{K}$

The value of $\gamma$can be calculated as:

$$\begin{gathered} \frac{{V_1}^{\gamma -1}}{{V_2}^{\gamma -1}}=\frac{T_2}{T_1} \\ (\gamma -1)\ln \left(\frac{1}{2}\right)=\frac{239K}{296K} \\ \gamma =1.309 \end{gathered}$$

$\gamma =\frac{{\text{C}}_{\text{p}}}{{\text{C}}_{\text{v}}}$

${\text{C}}_{\text{p}}-{\text{C}}_{\text{v}}=\text{R}$

$\text{1.309}=\frac{{\text{C}}_{\text{v}}\text{+ 8.314}}{{\text{C}}_{\text{v}}}$

$$\begin{gathered} {\text{C}}_{\text{v}}=26.9 \text{J/K} \text{mol} \\ {\text{C}}_{\text{p}}=35.2 \text{J/K} \text{mol} \end{gathered}$$