Gas Law Problem Help.?

Question

What volume will 300mL of gas at 20C and a pressure of 335mm Hg occupy if the temperature is reduced to -80C and the pressure increased to 760 mmHg?

Answer 1

See, first of all you need to convert your Celcius to Kelvin by the formula, 273+C=K. Therefore 20C becomes 293 K and -80C becomes 193. Now use this combined gas equation:
P1V1/T1=P2V2/T2 where P1 is the initial pressure, V1 is the initial volume and T1 is the initial temperature.
P1=335mm Hg
V1=300ml
T1=293K

whereas P2,V2 & T2 stand for final pressure, final volume (the one you have to find in this question), and final temperature.
P2=760mm Hg, V2=x, T2=193K

so substitute these values now,

335*300/293 = 760*x/193

x=335*300*193/760*293
x=87.1ml

Answer 2

OK, start with the Ideal Gas Law PV = nRT. In this case, it's more conveniently written in its other form, P1V1/n1T1 = P2V2/n2T2. Now n is held constant here, so P1V1/T1 = P2V2/T2. We're trying to solve for V2, the final volume, so we need to isolate it. Divide both sides by P2/T2: P1V1T2/T1P2 = V2. Now, plug in the numbers--but remember, you MUST convert the temperature to Kelvin first!!! 20 C is 293.15 K and -80 C is 193.15 K.

OK, NOW plug in the numbers. 335*300*193.15/(293.15*760) = V2.

Now, solve the equation: 87 mL = V2. This makes sense. Both lowering the temperature and increasing the pressure should decrease the volume. Since the pressure has almost doubled, and the temperature has been almost cut in half, you would expect the new volume to be a little more that 1/4 the original volume, and it is.

Answer 3

PV=nRT

(335mm)(.3 liters)/(.082)(293 K)=n=3.906 moles

V=(3.906)(.082)(193 K)/760mm =.081 liters