A high-pressure gas cylinder contains 40.0 L of toxic gas at a pressure of 1.20

Question

A high-pressure gas cylinder contains 40.0 L of toxic gas at a pressure of 1.20 107 N/m2 and a temperature of 23.0°C. Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperatures (-78.5°C), 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? Incorrect: Your answer is incorrect. N/m2 (b) What is the final pressure if one-tenth of the gas escapes? Incorrect: Your answer is incorrect. N/m2 (c) To what temperature must the tank be cooled to reduce the pressure to 1.00 atm (assuming the gas does not change phase and that there is no leakage during cooling)? Incorrect: Your answer is incorrect. K(d) Does cooling the tank appear to be a practical solution? No Yes Correct: Your answer is correct.

Explanation / Answer

Let V be the volume of the gas V1 = 40.0 L

Let the pressure of the gas be P1 = 1.20 * 107 N/m2

Let the temperature of the gas be T1 = 230 C

Then T1 = 23 + 273 = 296 K

Now we can easily solve the problem by using ideal gas equation or both of the Charles laws

a)Here the change in volume is negligible

so we can use Charles law at constant volume

               P1/T1 = P2/T2

Now the temperature of gas is T2 = - 78.50 C = -78.50 + 273 = 194.5 K

So final pressure P2 = P1 * T2/T1 = 1.20 * 107 * 194.5/296

                                    = 0.788 * 107 Pa

b)Now the volume changes but temperature does not change

         One tenth of the gas escapes so V2 = 0.9 V1

So                    P1 * V1 = P2 * V2

          1.20 *107 * V1 = P2 * 0.9 V1

                                  P2 = 1.20 * 107/0.9

                                       = 1.33 * 107 Pa

C)Now the volume again changes does not change but there is a change in pressure and there is a change in temperature

Let the final temperature be T2

Now final pressure be P2 = 1 atm = 1.013 * 105 Pa

                  P1/T1 = P2/T2

                        T2 = P2 * T1/P1 = 1.013 * 105 * 296/1.2 * 107

                             = 2.49 K

d)No. The final temperature is not enough to cool the gas completly

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