Question #cabd7

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Answer 1 · 2016-01-06T12:47:15

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

Using the Ideal Gas Law, $PV = nRT$ .

We can take the first scenario, with

$P = 1.03 atm$
$V = 2.20 L$
$R = 0.082057338 (atm L)/(mol K)$
$T = 320.15 K$ .

Which yields the following value for $n$ :

$n = (PV)/(RT) = (1.03 * 2.20)/(0.082057338 * 320.15)$

$n = 0.0862559 mol$

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

$P = 0.789 atm$
$T = 380.15K$ .

We get the following solution for $V$ :

$V = (nRT)/P = (0.0862559 * 0.082057338 * 380.15)/0.789$

$V = 3.4102357 L$

Using the 'combined gas law': $(P_1 V_1 )/ T_1 = (P_2 V_2) / T_2$ .

We can just plug in the values straight away!

$V_2 = (P_1 V_1)/(T_1) * (T_2)/(P_2)$

$V_2 = (1.03 * 2.20)/(320.15) * (380.15)/(0.789)$

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.