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.

Iwould you figure it out but you don't have a picture or in your question as to what the physical property you're looking for is