760 torr = 101.3 kPa = 14.7 psi = 1 atm

A machine has 30.0 L of helium kept at - 268ºC and a pressure of 207 torr. What would the new volume be if the helium is allowed to warm to 20.0ºC and 101.3 kPa?

(P1) (V1) / (T1) = (P2) (V2) / (T2)

Set up as: (P1) (V1) (T2) / (T1) (P2) = (V2)

P1=207torr

V1=30.0L

T1=-268°C

P2=101.3kPa (Same as 760torr) (use torr to make it easier though)

T2=20.0°C

Make sure all °C is in Kelvin

°C + 273 = Kelvin

Numbers are now:

P1=207torr

V1=30.0L

T1=5K

P2=760torr

T2=293K

Set them up as shown above

Should look like this:

(207torr) (30.0L) (293K) / (5K) (760torr) = V2

As long as you do them in that order you should get: 478.82L

You can also round to get 479L

Either one is correct.