91Ó°ÊÓ

The ideal gas obeys the ideal gas law when it undergoes change from the initial to the final state. The change in internal energy during this change of state depends on the change in temperature of the system.The ideal gas equation is given as-

$\mathbf{pV}\mathbf{=}\mathbf{nRT}\text{ }â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(\mathbf{1}\right)$

The change in internal energy is given as-

$\mathbf{Δ\pm «}\mathbf{=}\frac{3}{2}\mathbf{²Ô¸éΔ°Õ}\text{ }â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(\mathbf{2}\right)$

Here,Pis pressure, Vis volume, Tis temperature, Ris the gas constant and nis number of moles.

1. The number of moles of the ideal gas$n=3.00\text{mol}$
2. The initial pressure of ideal gas$p_1=20.0\text{atm}$
3. The initial volume of the ideal gas$V_1=1500\text{ c³¦}=1.5×{10}^{−3}{\text{″¾}}^{\text{3}}$
4. The pressure of the gas in state 2,$p_2=1.50\text{ }{p}_1$
5. The volume of the gas in state 2,$V_2=2.00V_1$
6. The pressure of the gas in state 3,$p_3=2.00p_1$
7. The volume of the gas in state 3,$V_3=0.50V_1$
8. The ideal gas constant,$R=8.31\text{J}/\text{mol}â‹\dots \text{K}$

(a)

From equation (1) the ideal gas law is given as-

$pV=nRT$

In state 1, the ideal gas equation can be written as-

$T_1=\frac{P_1{V}_1}{nR}$

For the given values of state parameters in state 1, the equation becomes-

$\begin{array}{c}{\text{T}}_1=\frac{(20.0×1.01×{10}^5\text{Pa})×(1.5×{10}^{−3}{\text{″¾}}^{\text{3}})}{\left(3.00\text{″¾ol}\right)×(8.31\text{ J}/\text{K.mol})}\\ =121.5\text{‿é}\end{array}$

(b)

As the state of the gas changes from 1 to 2, it obeys the gas law equation. Hence,

$\begin{array}{c}\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}\\ T_2=\frac{p_2{V}_2×T_1}{p_1{V}_1}\end{array}$

For the given values of state parameters in state 1 and 2, we have-

$\begin{array}{c}{T}_2=\frac{1.50p_1×2.00V_1×121.5\text{‿é}}{p_1{V}_1}\\ =1.50×2.00×121.5\\ =364.5\text{‿é}\end{array}$

(c)

As the state changes from 2 to 3, using the gas law for the change in state of gas from state 1 to state 3-

$\frac{p_1{V}_1}{T_1}=\frac{p_3{V}_3}{T_3}$

$T_3=\frac{P_3{V}_{3}xT1}{p_1{V}_1}$

For the given values of state parameters in state 1 and 3, we have-

$\begin{array}{c}{T}_3=\frac{2.00p_1×0.50V_1×121.5}{p_1{V}_1}\\ =2.00×0.50×121.5\text{‿é}\\ =121.5\text{‿é}\end{array}$

Thus, we find that the temperature of the gas in state 1 is the same as the temperature of the gas in state 3.

Next, the change in internal energy of the gas during the change from state 1 to state 3 is calculated as

$\mathrm{Δ}U=\frac{3}{2}nR\mathrm{Δ}T$

For $\mathrm{Δ}T=0$, the above equation becomes-

$\begin{array}{c}\mathrm{Δ}U=\frac{3}{2}nR×0\\ =0\end{array}$

Thus, there is no net change in internal energy of the gas.

The temperatures in state 1 and 2 are$121.5\text{ ‿é and364.5‿é}$ respectively. No, net change of internal energy change occurs between state 1 and 3.