A high-pressure gas cylinder contains 50.0 L of toxic gas at a pressure of 1.40 x 107 N/m2 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 1.0 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?
According to the ideal gas law,
Here, P is the pressure of the gas, V is the volume occupied by the gas, N is the number of molecules in the gas, k is the Boltzmann constant, and T is the absolute temperature. Let P0 be the initial pressure, V0 be the initial volume occupied by the gas, N0 be the number of molecules present initially, T0 be the initial temperature, Pf be the final pressure, Vf be the final volume, Nf be the final number of molecules and Tf be the final temperature. Then you can write,
First, you have to identify the given quantities.
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:
Therefore, the final pressure is 9.16 x 106 N/m2.
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.
Now the equation (2) can be modified as:
Therefore, the final pressure becomes 8.25 x 106 N/m2 when one-tenth of the gas escapes.
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 1.00 atm.
Modify equation (2) by removing the constant terms and substituting the given values.
So, the final temperature is 2.16 K
Cooling the tank to a temperature 2.16 K is practically impossible. Because that temperature is very low. So, it can’t be a practical solution.
Therefore, cooling is not a practical solution.
Question: Air in human lungs has a temperature of 37.0ºC and a saturation vapor density of 44.0 g/m3. (a) If 2.00 L of air is exhaled and very dry air inhaled, what is the maximum loss of water vapor by the person? (b) Calculate the partial pressure of water vapor having this density, and compare it with the vapor pressure of 6.31×103 N/m2.
Question: Atmospheric pressure atop Mt. Everest is 3.30×104 N/m2. (a) What is the partial pressure of oxygen there if it is 20.9% of the air? (b) What percent oxygen should a mountain climber breathe so that its partial pressure is the same as at sea level, where atmospheric pressure is 1.01×105 N/m2? (c) One of the most severe problems for those climbing very high mountains is the extreme drying of breathing passages. Why does this drying occur?
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