Question

If the initial pressure of an ideal gas at a temperature of 346 K is 0.946 atm, what is the pressure of the gas at 3.20 x 10^2 K?

Answer 1

Answer: 0.875 atm

Step-by-step explanation:

We can use the combined gas law: P1V1/T1 = P2V2/T2, where the subscripts designated initial (1) and final (2) conditions. We want P2, so rearrange:

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

Note how I've expressed the volume and temperature variables: as ratios. This helps visualize what to expect, and makes cancelling units easier and safer. We are not given a volume, so lets assume a convenient number, 1 liter. It remains constant, we have to assume, so the (V1/V2) cancels to 1. Temperatures must always be in Kelvin.

P2 = (0.946atm)*(1)*(320K/346K)

P2 = 0.875 atm