A high-pressure gas cylinder contains 60.0 L of toxic gas at a pressure of 1.31

Question

A high-pressure gas cylinder contains 60.0 L of toxic gas at a pressure of 1.31 ? 107 N/m2 and a temperature of 22.0°C. Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature (?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 in pascals, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? Pa

(b) What is the final pressure in pascals if one-tenth of the gas escapes? Pa

(c) To what temperature in kelvins 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)? K

(d) Does cooling the tank appear to be a practical solution?

Explanation / Answer

From idealgas equation we have

PV = nRT

(1.31 *(10^7) * 60*(10^-3) = n* 8.314 *(273 + 22)

n= 320.472 moles

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a) for constant volume

P1/ T1 = P2/T2
(1.31 *(10^7)) / (273 + 22) = P2/ (273 -78.5)
P2 = 8.637*10^6 Pa

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b)
If one tenth escapes,

n = 0.9 * 320.472

n= 288.425 moles

so, P*V = n*R*T
P* 60 *(10^-3) = 288.425 * 8.314 *(273 - 78.5)

P = 7.773*10^6 N/m^2

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c)
for P = 1 atm = 101325 N/m^2
so, P1/T1 = P2/T2
(1.31 * (10^7))/(273 + 22) = 101325/ (273 + T2)
T2 = - 270.72 K

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d)
T2 = 273 - 270.72 = 2.28 C

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Comment incase any doubt..goodluck

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