# A 3.00 liter sample of gas is at 288 K and 1.00 atm. If the pressure of the gas is increased to 2.00 atm and its volume is decreased to 1.50 liters, what will be the Kelvin temperature of the sample?

Jan 2, 2016

When you are given this much information in the context of a generic, unnamed gas, it's a good idea to consider the Ideal Gas Law:

$P V = n R T$

Before we proceed any further, notice what happened. It's quite funny, actually.

The pressure doubled, and the volume halved, thus numerically cancelling each other out. Therefore, we can expect the final temperature be the same as the initial temperature!

Still, proceed to see the calculations below, just to get an idea of how to do the actual calculation.

Note that the number of $\text{mol}$s doesn't change, and either does $R$, the universal gas constant.

Since the final temperature is not known, but the number of $\text{mol}$s is not known at all, we should solve for $n$ first.

$n = \frac{{P}_{i} {V}_{i}}{R {T}_{i}}$

$= \left(\left(\text{1.00" cancel"atm")("3.00" cancel"L"))/(("0.082057" cancel"L"cdotcancel"atm""/mol"cdotcancel"K")("288" cancel"K}\right)\right)$

$=$$\text{0.1269 mol gas}$

That will stay constant, because we are implicitly assuming a closed system for simplicity.

Now, let's solve for ${T}_{f}$.

${T}_{f} = \frac{{P}_{f} {V}_{f}}{n R}$

$= \left(\left(\text{2.00" cancel"atm")("1.50" cancel"L"))/(("0.1269" cancel"mol gas")("0.082057" cancel"L"cdotcancel"atm""/"cancel"mol"cdot"K}\right)\right)$

$=$$\textcolor{b l u e}{\text{288 K}}$

Yep! The same temperature.