A high-pressure gas cylinder contains 50.0 L of toxic gas at...

A high-pressure gas cylinder contains 50.0 L of toxic gas at a pressure of $1.40 \times 10^{7} \mathrm{N} / \mathrm{m}^{2}$ and a temperature of $25.0^{\circ} \mathrm{C} .$ Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature $\left(-78.5^{\circ} \mathrm{C}\right)$ 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 1.00 atm (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?

Key Concepts

Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, is a fundamental equation in thermodynamics that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas through the gas constant (R). It serves as the primary tool for predicting how changes in one state variable affect the others under ideal conditions, and is particularly useful in problems involving controlled environments, such as gas cylinders, where one or more parameters remain constant.

Temperature–Pressure Relationship (Gay-Lussac's Law)

Under conditions of constant volume, the pressure of a gas is directly proportional to its absolute (Kelvin) temperature, a principle known as Gay-Lussac’s law. This concept allows one to determine changes in pressure when a gas is cooled or heated, assuming no gas escapes and the container’s volume does not change.

Effects of Gas Leakage on System Variables

When a controlled system such as a gas cylinder experiences a leak, the number of moles of the gas (n) decreases, leading to a corresponding decrease in pressure if the temperature and volume remain constant. Understanding how the loss of a fraction of the gas affects the overall pressure is crucial when applying the Ideal Gas Law to real-world scenarios, particularly in safety-critical applications.

Phase Change Considerations in Gas Cooling

In thermodynamic processes, it is important to consider whether a substance will remain in the gaseous phase when subjected to changes in temperature and pressure. In problems where the gas is cooled, ensuring that the gas does not condense or transition into another phase is essential for the application of the Ideal Gas Law, as phase changes introduce additional complexities not accounted for by the simple ideal gas relationship.

Transcript

00:01 So for part a, from the ideal gas law, we know that p sub 1 v.

00:04 1 divided by the number of molecules or atoms times boltzman's constant times t sub 1, this would be equalling p sub 2 v sub 2 divided by n sub 2 k, t sub 2.

00:27 And so we can then see that for one, subscript one and two indicate the quantities at the initial and final time respectively.

00:39 We know the volume didn't change so we can eliminate this term from both sides.

00:44 We can eliminate bolt spin constant from both sides.

00:48 And we know that the number of molecules for the final and the initial step are approximately the same.

00:55 So we can eliminate n sub 1 and n sub 2 as well.

00:59 And so we finally have that p .2 is equaling p sub 1.

01:06 The pressure sub 2 equals the pressure sub 1, multiplied by the temperature at 2 minus, rather divided by the temperature at 1.

01:14 And so we can then say that the pressure at 2 would be equaling the pressure at 1, 1 .40, times 10 to the 7th newton's per square meter this would be multiplied by t 2 negative 78 .5 plus 273 .15 so we have to convert from celsius to kelvin by adding 273 .15 we do the exact same thing for t sub 1 so 25 plus 273 .15 and and we can say that then pressure sub 2, 9 .14 times 10 to the 6th, newtons per square meter.

02:03 So this would be the final pressure for part a.

02:08 And then for part b, we know that here, 1 tenth of the gas escapes.

02:17 So the number of atoms will also reduce by one tenth.

02:21 So n sub 2 over n sub 1, we know is going to be equal to 1 minus 1 tenth, or this would be equal to 0 .9.

02:33 And so n sub 2 is equaling 0 .9 n sub 1.

02:39 And so we can then say that p sub 2 is going to be equal to p sub 1, multiplied by temperature sub 2 divided by temperature sub 1.

02:50 And we add another factor n sub 2 divided by n sub 1...

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