An ideal gas at a pressure of 152 kPa is contained in a bulb
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Question

An ideal gas at a pressure of 152 kPa is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of 0.800 L as shown here. When the stopcock is opened, the gas expands into the empty bulb. If the temperature is held constant during this process and the final pressure is 92.66 kPa, what is the volume of the bulb that was originally filled with gas?

Accepted Answer

Hi everyone here we have a question telling us that two bulbs are connected as shown below. The left bulb initially contains 1.76 atmospheres of an ideal gas. The stop cock is opened and the ideal gas is allowed to expand at a constant temperature into the right bulb with a volume of 0.500 liters, calculate the volume of the left bulb if the final pressure of the ideal gas is 0.8 atmospheres. So we're going to use Boyle's law, which is pressure one Times vol one equals pressure too, times volume two. So our pressure one is going to equal 1. atmospheres. Our Volume one is going to equal x. Our pressure to Is going to equal 0.88 atmospheres. Our volume two is going to equal X plus 0.500 liters. So we are going to take one .76 Atmospheres times X Equals 0.88 atmospheres Times x plus 0.500 L. And that is going to equal 1.76 atmospheres times eggs Equals 0.88 atmospheres Times x plus 0. atmospheres Times Leaders. So zero 0. atmospheres times X Equals 0. atmospheres. Times Leaders. So x equals v one which equals 0. L. And that is our final answer. Thank you for watching. Bye

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