91Ó°ÊÓ

According to the ideal gas law,

PV = NkT……………..(1)

Here, is the pressure of the gas, is the volume occupied by the gas, is the number of molecules in the gas, is the Boltzmann constant, and is the absolute temperature. Let be the initial pressure, be the initial volume occupied by the gas, be the number of molecules present initially, be the initial temperature, be the final pressure, be the final volume, be the final number of molecules and be the final temperature. Then you can write,

$\frac{{\mathbf{P}}_{\mathbf{0}}{\mathbf{V}}_{\mathbf{0}}}{{\mathbf{P}}_{\mathbf{f}}{\mathbf{V}}_{\mathbf{f}}}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{0}}{\mathbf{T}}_{\mathbf{0}}}{{\mathbf{N}}_{\mathbf{f}}{\mathbf{T}}_{\mathbf{f}}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(2)

First, you have to identify the given quantities.

$$\begin{gathered} P_0=1.40×{10}^7 \\ {T}_0=25°\text{C}+273.15 \\ =298.15\text{K} \end{gathered}$$

$$\begin{gathered} {\text{T}}_f=-78.5°\text{C}+273.15 \\ =195.15\text{K} \\ {\text{V}}_0=\frac{50.0\text{L}}{1000{\text{L/m}}^3} \\ =0.05{\text{m}}^3 \end{gathered}$$

It is given that the amount of gas leaked out is negligible. So, the number of molecules remains constant. The volume of the gas also remains constant. Then equation (2) can be modified as:

$$\begin{gathered} \frac{P_0}{P_f}=\frac{T_0}{T_f} \\ {P}_f=\frac{P_0{T}_f}{T_0} \\ {P}_f=\frac{1.40×{10}^7{\text{N/m}}^2×195.15\text{K}}{298.15\text{K}} \\ {P}_f=9.16×{10}^6{\text{N/m}}^2 \end{gathered}$$

Therefore, the final pressure is$9.16×{10}^{6}N/m^2$ .

The volume remains constant, but the number of molecules in the final stage will be 0.9 times that of the initial number of molecules. Because one-tenth of the molecules got away.

.$$\begin{gathered} N_f=\frac{9}{10}{N}_0 \\ \frac{N_f}{N_0}=\frac{9}{10} \end{gathered}$$

Now the equation (2) can be modified as:

$$\begin{gathered} \frac{P_0}{P_f}=\frac{N_0{T}_0}{N_f{T}_f} \\ {P}_f=\frac{P_0{T}_f}{T_0}\frac{N_f}{N_0} \\ {P}_f=\frac{1.40×{10}^7{\text{N/m}}^2×195.15\text{K}}{298.15\text{K}}×\frac{9}{10} \\ {P}_f=8.25×{10}^6{\text{N/m}}^2 \end{gathered}$$

Therefore, the final pressure becomes when one-tenth of the gas escapes.

The gas has not leaked and phase change has not occurred. So, the number of molecules and volume remains constant. Here you need to calculate the final temperature if the final pressure is .

$$\begin{gathered} {\text{P}}_f=1.00\text{atm}×1.013×{10}^5{\text{N/m}}^2{\text{atm}}^{-1} \\ =1.013×{10}^5{\text{N/m}}^2 \end{gathered}$$

Modify equation (2) by removing the constant terms and substituting the given values.

$$\begin{gathered} \frac{P_0}{P_f}=\frac{T_0}{T_f} \\ {T}_f=\frac{T_0{P}_f}{P_0} \\ {T}_f=\frac{298.15\text{K}×1.013×{10}^5{\text{N/m}}^2}{1.40×{10}^7{\text{N/m}}^2} \\ {T}_f=2.16\text{K} \end{gathered}$$

So, the final temperature is 2.16 k .

Cooling the tank to a temperature is 2.16 K practically impossible. Because that temperature is very low. So, it can’t be a practical solution.

Therefore, cooling is not a practical solution.