Question

The gas inside of a container exerts `5 Pa` of pressure and is at a temperature of `120 ^o C`. If the temperature of the gas changes to `320 ^oK` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

OK, there is something odd here, which might explain the delayed response... you quote one temp as `120^@C` and the second as `320K`

Explanation:

Assuming that is what you meant we just need to convert the degrees C into K (so we have consistent units) and it should work.

`120^@C = 120 + 273 = 393`K

OK, so it is at a pressure, p = 5Pa and cools from 393 to 320K. As `pV = nRT` and V, n & R are all constant (so cancel) we can compare the two scenarios as a pair of ratios:

`p_1/p_2 = T_1/T_2` and substitute in the data to solve for `p_2`

`5/p_2 = 393/320`

So `p_2 = (5 xx 320)/393`

Thus `p_2 = 4.07`Pa which I’d probably quote as `p_2 = 4.1`Pa given the sig. figs. given in the data.