Chemistry Help Please

Question

Consider 6.07 L of a gas at 303.9 mmHg and 20.0 ºC . If the container is compressed to 2.41 L and the temperature is increased to 55.8 ºC , what is the new pressure inside the container? Assume no change in the amount of gas inside the cylinder. Report the pressure to the tenths of a mmHg without units.

Answer 1

Initially:

V1 = 6.07 L

P1 = 303.9 mmHg

T1 = 20.0 deg C

Finally:

V2 = 2.41 L

P2 = ?

T2 = 55.8 deg C

***IMPORTANT***

Units must match with each other (example if V1 is liters then V2 must be liters). This means our answer for P2 will come out in mmHg because those are the units of P1.

Temperature MUST BE Kelvin! T1 and T2 need to be converted before any calculations.

Celsius + 273 = Kelvin

T1 —> 20.0 + 273 = 293.0 K

T2 —> 55.8 + 273 = 328.8 K

Choose one of our gas laws that contains P, V, and T: Combined gas law

P1V1/T1 = P2V2/T2

Plug in our knowns:

(303.9*6.07)/(293.0) = (P2 * 2.41)/(328.8)

Cross multiplying is one method you can use to start every gas law problem regardless of which variable is missing. The denominator on one side matches with the numerator of the other side to get rid of all denominators:

303.9 * 6.07 * 328.8 = P2 * 2.41 * 293.0

multiply

606,528.4824 = P2 * 706.13

divide by 706.13

P2 = 606,528.4824 / 706.13 = 858.9 mmHg

(If your teacher has you use 273.15 for kelvin, you might have a slightly different answer)