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?

1 Answer
Jan 8, 2018

Answer:

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.