# Question #cabd7

Jan 6, 2016

They should give the same answer of $V = 3.41 L$, assuming that the amount of gas (i.e., moles) is kept constant.

#### Explanation:

Using the Ideal Gas Law, $P V = n R T$.

We can take the first scenario, with

$P = 1.03 a t m$
$V = 2.20 L$
$R = 0.082057338 \frac{a t m L}{m o l K}$
$T = 320.15 K$.

Which yields the following value for $n$:

$n = \frac{P V}{R T} = \frac{1.03 \cdot 2.20}{0.082057338 \cdot 320.15}$

$n = 0.0862559 m o l$

Using that value for $n$ on the latter situation, with

$P = 0.789 a t m$
$T = 380.15 K$.

We get the following solution for $V$:

$V = \frac{n R T}{P} = \frac{0.0862559 \cdot 0.082057338 \cdot 380.15}{0.789}$

$V = 3.4102357 L$

Using the 'combined gas law': $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$.

We can just plug in the values straight away!

${V}_{2} = \frac{{P}_{1} {V}_{1}}{{T}_{1}} \cdot \frac{{T}_{2}}{{P}_{2}}$

${V}_{2} = \frac{1.03 \cdot 2.20}{320.15} \cdot \frac{380.15}{0.789}$

Which, allegedly equals,

${V}_{2} = 3.4102357 L$, which is as we expected.

I would say the latter method is much faster, however, it's not as 'general' as the Ideal Gas Law.