Im stuck on this physics question. A sample of helium gas etc..?

Question

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperture of 29 celsics. What is the new temp (C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Answer 1

The volume halved while the pressure quadrupled, so the temperature doubled (PV doubled).   T = 2 (29+273) Kelvin = 58+273 C.

Answer 2

I came up with 58 degrees Celsius

P1*V1    P2*V2

-------- = -----------

T1*n1      T2*n2

solve for T2

T2 =  P2*V2*T1

           --------------

                 P1*V1

T2= 3.2*.09L*29c

          ------------------

            .800*.180L

Answer 3

1st have to assume the ideal gas law pv = nRt and conver the temps to absolute so the law applies

The idea here is to use ratios so you can not worry about n which is unkown and does not change anyway

pv/t = pv/t

0.8 * 0.18/302 = 3.2*.09/t

t = 3.2*0.09/(0.8*.18/302)

= 604 K

=331 C

This makes sense since you halved the volume and quadroupled the pressure, temp should rise alot