If a sample of H2 has a volume of 22.4 L at 273.0 K and 760.0 mmHg, what will the volume be if the pressure decreases to 650.0 mmHg and the temperature increases to 320.0K?

30.7L

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 22.4L

T1 (initial temperature) = 273K

P1 (initial pressure) = 760mmHg

P2 (final pressure) = 650mmHg

T2 (final temperature) = 320K

V2 (final volume) = ?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume of the gas can be obtained as follow:

P1V1/T1 = P2V2/T2

760x22.4/273 = 650xV2/320

Cross multiply to express in linear form

273 x 650 x V2 = 760 x 22.4 x 320

Divide both side by 273 x 650

V2 = (760 x 22.4 x 320) / (273 x 650)

V2 = 30.7L

Therefore, the new volume of the gas will be 30.7L

the serious answer:

final volume or V2 = 30.7 liters

Explanation:

Considering that the values that this problem gives you are: the value of V1 or initial volume of the gas, T1 and T2 which are initial and final temperature, and as for the pressure it gives you P1 and P2, that is, the values also of initial pressure and finally.

So up to here we have all the values, except for the final volume, that is, when that gas changed its temperature and pressure ... what V2 or volume did it reach?

That is why we will use the following equation:

P1 x V1 / T1 = p2 x v2 / T2

We clear V2 from this equation, so as the final equation we would have:

V2 = (760X22,4X320) / (273X650) = 30.7 liters