91影视

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-

$\mathbf{W}\mathbf{=}\mathbf{笔螖痴}$

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

1. Theconstant pressure is $P=250\text{鈥塒补}$
2. The initial volume of the ideal gasis$V_i=0.80{\text{鈥尘}}^3$
3. The final volume of the ideal gasis$V_f=0.20{\text{鈥尘}}^3$
4. The initial temperature of thegas is$T_i=360\text{鈥块}$
5. The heat lost by the gas is $Q=鈭�210\text{鈥塉}$

The work done at constant pressure is

$$\begin{gathered} W=P螖V \\ W=P(V_f鈭�V_i) \end{gathered}$$

For the given values, we have-

$W=250\text{鈥塒补}(0.20{\text{鈥尘}}^{\text{3}}鈭�0.80{\text{鈥尘}}^{\text{3}})$

$W=鈭�150\text{鈥塉}$

The heat lost by the gas is $Q=鈭�210\text{鈥塉}$. From the first law of Thermodynamics-,

$$\begin{gathered} 螖E_{\mathrm{int}}=Q鈭�W \\ 螖E_{\mathrm{int}}=鈭�210\text{鈥塉}鈭�(鈭�150\text{鈥塉}) \end{gathered}$$

$螖E_{\mathrm{int}}=鈭�60\text{鈥塉}$

(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.

$V鈭�T$

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-

$\begin{array}{c}{T}_f=\frac{0.20{\text{鈥尘}}^3}{0.80{\text{鈥尘}}^3}脳360\text{鈥块}\\ =90\text{鈥块}\end{array}$

The decrease in internal energy of the gas is $60\text{鈥塉}$ and the final temperature of the gas is $90\text{鈥块}$.