Question Video: Cooling and Compression of a Cloud Physics • 9th Grade

A cloud has a volume of 3,560 mยณ when in air with a pressure of 106 kPa and a temperature of 290 K. The air temperature then drops to 275 K and the air pressure drops to 101 kPa, causing the cloud to compress, as shown in the diagram. What is the new volume of the cloud? Answer to the nearest cubic meter.

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Video Transcript

A cloud has a volume of 3,560 cubic meters when in air with a pressure of 106 kilopascals and a temperature of 290 kelvin. The air temperature then drops to 275 kelvin and the air pressure drops to 101 kilopascals, causing the cloud to compress, as shown in the diagram. What is the new volume of the cloud? Answer to the nearest cubic meter.

In this situation, our cloud starts out with a volume shown by this solid black outline. But then, as the temperature and the pressure of the cloud decrease, the cloud compresses to a smaller volume. If we call ๐‘‰ one the initial volume of the cloud and ๐‘‰ two its final volume, itโ€™s that volume ๐‘‰ two that we want to solve for. If we treat this cloud as an ideal gas, then we know that it follows the ideal gas law. Here, ๐‘ƒ is the gasโ€™s pressure, ๐‘‰ is its volume, ๐‘› is the number of moles of the gas, ๐‘… is a constant called the gas constant, and ๐‘‡ is the gas temperature.

If we rearrange this equation somewhat so that ๐‘ƒ, ๐‘‰, and ๐‘‡ are on one side and ๐‘› and ๐‘… are on the other, we can notice that in our scenario, ๐‘› is a constant โ€” that is, the number of moles of the gas in this cloud is constant โ€” and ๐‘… is always a constant, so ๐‘› times ๐‘… is itself a constant value.

This means if we take the pressure of the gas at one moment, call it ๐‘ƒ one, and multiply it by the volume of the gas at that same moment, call that ๐‘‰ one, and divide all that by the temperature of the gas at the same time ๐‘‡ one, then thatโ€™s equal to the pressure times the volume divided by the temperature at some other moment. Weโ€™ll signify this with a subscript two. In this equation, we can think of the values on the left-hand side as the pressure, volume, and temperature of our cloud before its volume changes and then the values on the right as those values after its volume changes.

Thinking of it this way, we know that itโ€™s ๐‘‰ two that we want to solve for. And we can start to do this by multiplying both sides of the equation by ๐‘‡ two divided by ๐‘ƒ two. We do this so that the temperature ๐‘‡ two will cancel out on the right as well the pressure ๐‘ƒ two. This gives us an equation where the volume ๐‘‰ two is the subject. Itโ€™s equal to ๐‘‰ one, the initial volume of the cloud, times this ratio of pressures, ๐‘ƒ one to ๐‘ƒ two, times this ratio of temperatures, ๐‘‡ two to ๐‘‡ one.

In our problem statement, weโ€™re given a number of these values. For example, the initial volume of the cloud, weโ€™ll call it ๐‘‰ one, is 3,560 cubic meters. The initial cloud pressure, weโ€™ll call it ๐‘ƒ one, is 106 kilopascals, while the final pressure of the gas, weโ€™ll call it ๐‘ƒ two, is 101 kilopascals. Weโ€™re also told that the initial temperature of the cloud is 290 kelvin; weโ€™ll call that ๐‘‡ one. And the final temperature of the cloud is 275 kelvin; weโ€™ll call that ๐‘‡ two. Weโ€™re just about ready to substitute into our equation and calculate ๐‘‰ two.

Before we do that, letโ€™s clear some space on screen, and with our values input into the equation, letโ€™s look for a moment at the units. In terms of pressure, we have units of kilopascals in numerator and denominator. So, these units will cancel out in our calculation. The same thing happens to our units of kelvin for temperature. In the end, weโ€™ll have an answer in units of cubic meters, a volume, just as we want. Knowing this, when we calculate ๐‘‰ two and round our result to the nearest cubic meter, we get 3,543 cubic meters. This is the new volume of the cloud after it has been compressed. Note that this final volume is only slightly less than the initial volume ๐‘‰ one.

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