A gas has a pressure of 1.5 atm at a temperature of 30 degrees C. If the gas is cooled to 10 degrees C, what is the new pressure?

Answer 1

Hello, Jake,

When there is no change in the moles of a gas between two states, we can use a simplified version of the deal gas law:

P1V1/T1 = P2V2/T2

This is basically PV=nRT for two conditions (1 and 2) set into one equation that results in both n, the moles, and R, the gas constant, to cancel.

We want to find the final pressure, P2. Rearrange the equation to solve for P2:

P2 = P1(V1/V2)(T2/T1)

Note that the final pressure is the product of the initial pressure P1, times ratios of the beginning and final values for both the temperature and volume.

We're in luck again, because there is no mention of a change in volume, so we can set V1 = V2, and that factor then cancels. (V1/V2)=1.

We are left with P2 = P1(T2/T1). This means if we decrease the temperature (PT2<T1) we'll see an decrease in pressure, and vice versa.

Looks simeple, but don't forget we need degrees Kelvin for the gas laws. Add 273.15 to both Centigrade temperatures for to get degrees Kelvin.

Set up a table as you work through this problem. Then enter the data for P1, T1, and T2, and solve for P2. We expect a decrease in temperature, as well as final units of only pressure. Since we changed to Kelvin, the ratio of the two temperatures is greatly decreased, so we should expect only a small decrease in pressure.

I obtain P2 = 1.4 atm. Lower than initial and units of pressure, as we predicted.

Please ask if you have a question on this explanation,

Bob