Problem with a gas law equation?

Question

a sample of gas occupies 3.75 liters ar 16.91 lb/sq in and 25 degrees Celsius. at what temperature in fahrenheit will it occupy 4 liters at 1 atm?

my problem is finding one of the variables to solve the equation

i defined my variables here:
volume 1= 3.75 liters
volume 2= 4 liters
Temp1=25 degrees celsius+273= 298 kelvin
Temp2= (this is what we are finding)
pressure1= 1 atm
pressure2= ????? i dont understand how to find the second pressure

if someone can please show me how to find that second pressure variable that would be so helpful

also do i have to convert the 16.91 lb/sq in for the volume?

Answer 1

First, as your using psi together with litres and °C, you're better converting psi to Atmospheres.
16.91psi ÷ 14.7psi/atm = 1.15atm.
Now, Use the Combined Gas Law. (T = Kelvin (as you have done). (Your P1 is 1.15atm, P2 = 1 atm).

P1 x V1 x T2 = P2 x V2 x T1
1.15atm x 3.75L x T2 = 1atm x 4L x 298K
T2 = (1 x 4 x 298) ÷ (1.15 x 3.75)
T2 = 1,192 ÷ 4.313
Final Temp. = 276.4K (-273) = 3.4°C
(3.4°C x 1.8) + 32 = 38.12°F.

(Steve, No offence but, 492 R (32 °F) = 273 K (0°C) and, 1 atm is considered to be 14.696 (14.7 psi)

Answer 2

P1V1/T1 = P2V2/T2

T2 = P2V2T1/P1V1

V1 = 3.75L V2 = 4L T = 298K = 536R (Rankin is the absolute temperature scale that uses degF) P1 = 1 atm = 14.8psi

T2 = (14.8psi)(4L)(536R)/(16.91psi)(3.75L) = 500 degR

500 - 459 = 41F (the answer)

459 in degF for Rankin temperatures is like 273 for Kelvin temperatures.