A sample of helium gas has a volume of 788 L at a pressure of 76 cmHg and a temperature of 28° C. When the volume is increased to 1290 L, what is the temperature of the helium?

Question

Please explain how you got the answer, thank you

Answer 1

Since no change in pressure is mentioned, we can assume it's constant. So, we could use the combined gas law, (P₁V₁)/T₁ = (P₂V₂)/T₂. Since P is constant it drops out of the equation, giving Charles' Law,

V₁/T₁ = V₂/T₂. Since the T must be in Kelvins, we must convert the ^oC to K before doing the math:

K = ^oC + 273. So then T₁ = 28 + 273 = 301 K

Next, rearrange the equation to solve for T₂:

T₂ = (V₂T₁)/V₁ Put in the given values & solve. (1290 L)(301 K)/(788 L) = 492.8 = 493 K

If you are required to record the T in Celsius, convert it back by subtracting from 273.