Question

A sample of a gas has a volume of 426mL at 330 K. What temperature is necessary for the gas to have a volume of 475 mL? Remember to convert mL to L. Round to the nearest whole number.

Answer 1

V1 = 426 mL = 0.426 L

V2 = 475 mL = 0.475 L

T1 = 330 K

T2 = unknown

Solution:

Apply Charles' Law:

V₁/T₁ = V₂/T₂

Therefore,

T₂ = V₂*T₁ / V₁ = (0.475 L)(330 K)/(0.426 L) = 368 K

T₂ = 368 K.

Answer: T₂ = 368 K.