During a compression at a constant pressure of $\mathbf{250}\mathbf{}\mathbf{\text{\hspace{0.17em}Pa}}$, the volume of an ideal gas decreases from $\mathbf{0}\mathbf{.80}{\mathbf{\text{\hspace{0.17em}m}}}^{\mathbf{3}}$to$\mathbf{0}\mathbf{.20}{\mathbf{\text{\hspace{0.17em}m}}}^{\mathbf{\text{3}}}$ . The initial temperature is $\mathbf{360}\mathbf{\text{\hspace{0.17em}K}}$, and the gas loses $\mathbf{210}\mathbf{\text{\hspace{0.17em}J}}\mathbf{}$as heat.

1. What is the change in internal energy of gas?
2. What is the final temperature of the gas?

If during a process, the pressure of the gas remains constant, then the process is known as isobaric process. The work done $\left(\mathbf{W}\right)$by the system during an isobaric process, is given as-

$\mathbf{W}\mathbf{=}\mathbf{P\Delta V}$

Here P is the pressure and $\mathbf{\Delta V}$is the change in volume.

1. Theconstant pressure is $P=250\text{\hspace{0.17em}Pa}$
2. The initial volume of the ideal gasis$V_i=0.80{\text{\hspace{0.17em}m}}^3$
3. The final volume of the ideal gasis$V_f=0.20{\text{\hspace{0.17em}m}}^3$
4. The initial temperature of thegas is$T_i=360\text{\hspace{0.17em}K}$
5. The heat lost by the gas is $Q=-210\text{\hspace{0.17em}J}$

The work done at constant pressure is

$$\begin{gathered} W=P\Delta V \\ W=P(V_f-V_i) \end{gathered}$$

For the given values, we have-

$W=250\text{\hspace{0.17em}Pa}(0.20{\text{\hspace{0.17em}m}}^{\text{3}}-0.80{\text{\hspace{0.17em}m}}^{\text{3}})$

$W=-150\text{\hspace{0.17em}J}$

The heat lost by the gas is $Q=-210\text{\hspace{0.17em}J}$. From the first law of Thermodynamics-,

$$\begin{gathered} \Delta E_{\mathrm{int}}=Q-W \\ \Delta E_{\mathrm{int}}=-210\text{\hspace{0.17em}J}-(-150\text{\hspace{0.17em}J}) \end{gathered}$$

$\Delta E_{\mathrm{int}}=-60\text{\hspace{0.17em}J}$

(b) The final temperature of the gas:

According to Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant temperature.

$V\propto T$

We can apply this law for initial and final volume and temperature and take their ratio as

$\begin{array}{l}\frac{V_f}{V_i}=\frac{T_f}{T_i}\\ T_f=\frac{V_f}{V_i}{T}_i\end{array}$

For the given values, we have-

$\begin{array}{c}{T}_f=\frac{0.20{\text{\hspace{0.17em}m}}^3}{0.80{\text{\hspace{0.17em}m}}^3}\times 360\text{\hspace{0.17em}K}\\ =90\text{\hspace{0.17em}K}\end{array}$

The decrease in internal energy of the gas is $60\text{\hspace{0.17em}J}$ and the final temperature of the gas is $90\text{\hspace{0.17em}K}$.