# 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?

Jan 8, 2018

OK, there is something odd here, which might explain the delayed response... you quote one temp as ${120}^{\circ} C$ and the second as $320 K$

#### 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.

${120}^{\circ} C = 120 + 273 = 393$K

OK, so it is at a pressure, p = 5Pa and cools from 393 to 320K. As $p V = n R T$ 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 ${p}_{2}$

$\frac{5}{p} _ 2 = \frac{393}{320}$

So ${p}_{2} = \frac{5 \times 320}{393}$

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.