91Ó°ÊÓ

$$\begin{gathered} P_b=7.5\text{kPa}=7500\text{Pa} \\ {P}_{ac}=2.5\text{ k±Ê²¹}=2500\text{ P²¹} \end{gathered}$$

1. At point a, $T=200\text{‿é}$
2. At point a, $V=1{\text{″¾}}^3$
3. At point b,$V=3{\text{″¾}}^3$
   

Find thenumber of moles and the temperature of the gas at points b and c present in the sample using the gas law. Net energy added can be found from the net work done by the gas using the given graph.

Formula is as follow:

$p_i{v}_i=nR{T}_i$

Here, p is pressure, v is volume, T is temperature, R is universal gas constant and n is number of moles.

According to the gas law,

$$\begin{gathered} PV=nRT \\ n=\frac{PV}{RT} \end{gathered}$$

From the given graph, at point a,

$$\begin{gathered} n=\frac{2500\left(1\right)}{8.314\left(200\right)} \\ n=1.5\text{″¾ol} \end{gathered}$$

Hence, the number of moles of gas present in the sample are .$\text{1.5mol}$

By gas law,

$\frac{P_b{V}_b}{P_a{V}_a}=\frac{T_b}{T_a}$

$$\begin{gathered} T_b=\frac{P_b{V}_b}{P_a{V}_a}{T}_a \\ {T}_b=\frac{7500\left(3\right)}{2500\left(1\right)}\left(200\right) \end{gathered}$$

$T_b=1.8×{10}^3\text{K}$

Hence, the temperature of the gas at point b is $1.8×{10}^3\text{K}$.

Similarly,

$$\begin{gathered} T_c=\frac{P_c{V}_c}{P_a{V}_a}{T}_a \\ {T}_c=\frac{2500\left(3\right)}{2500\left(1\right)}\left(200\right) \end{gathered}$$

$T_c=600=6.0×{10}^2\text{K}$

Hence,the temperature of the gas at point c is$6.0×{10}^2\text{K}$.

First law of thermodynamics gives,

$△E=Q−W$

For isothermal process,

$$\begin{gathered} Q=W \\ Q=\text{Areaunderthecurve} \\ Q=\frac{1}{2}\left(\text{ac}\right)\left(\text{bc}\right) \end{gathered}$$

Substitute the values as:

$Q=\frac{1}{2}\left(2\right)(7500−2500)$

$Q=5000=5.0×{10}^3\text{J}$

Hence, the net energy added to the gas as heat during the cycle is$5.0×{10}^3\text{J}.$

Therefore, the number of moles of gas present in the sample,temperature at a point, and heat added can be foundusing gas law.