A high-pressure gas cylinder contains 40.0 L of toxic gas at a pressure of 1.10

Question

A high-pressure gas cylinder contains 40.0 L of toxic gas at a pressure of 1.10 107 N/m2 and a temperature of 27.0°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? 7181452 Correct: Your answer is correct. N/m2

(b) What is the final pressure if one-tenth of the gas escapes? 6463306 Correct: Your answer is correct. 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)? 2.06 Incorrect: Your answer is incorrect. K

(d) Does cooling the tank appear to be a practical solution? Yes Correct: Your answer is correct.

Explanation / Answer

a) Assuming the cylinder has a negligible volume shrinkage as it is cooled.
Apply the universal gas law ======> P1.V1/T1 = P2.V2/T2 and V1 = V2

Final pressure P2 = P1.T2 / (T1) (T in Kelvin)
P2 = (1.1*10^7 Pa)(273-78.50)K / (273+27) ========> P2 = 7131666.67 Pa

b) From PV = nRT .. .. R (constant) = PV/nT
So .. P1.V1 / (n T1) = P2.V2/(0.9nT2) (again V1 = V2)
P2 = P1*0.9*T2 / T1 = 6418500.0 Pa

c) P1/T1 = P2/T2 .. (VI = V2)(P2 = 1atm = 1.1*10^5 Pa)
T2 = P2.T1 / P1 =====> (1.1*10^5Pa x 300K) / 1.1*10^7Pa
T2 = 3.0 K

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