Question

Question: A high-pressure gas cylinder contains of toxic gas at a pressure of and a temperature of . Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature () to reduce the leak rate and pressure so that it can be safely repaired. (a) What is the final pressure in the tank, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? (b) What is the final pressure if one-tenth of the gas escapes? (c) To what temperature must the tank be cooled to reduce the pressure to (assuming the gas does not change phase and that there is no leakage during cooling)? (d) Does cooling the tank appear to be a practical solution?

Short answer

Answer:

1. The final pressure is $9.16\times {10}^{6}N/m^2$.
2. The final pressure when one-tenth of the gas escapes is $8.25\times {10}^{6}N/m^2$.
3. The final temperature is 2.16 k.
4. Cooling the tank is not a practical solution.

Step-by-step solution

State the ideal gas law.

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,

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

 Step 2: (a) Calculate the final pressure.

First, you have to identify the given quantities.

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

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

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:

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

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

(b) Calculate the final pressure when one-tenth of the gas escapes into the environment.

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:

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

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

(c) Calculate the final temperature.

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 .

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

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

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

So, the final temperature is 2.16 k .

(d) Describe whether cooling the tank appears to be a practical solution.

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.