When a sample of a gas at a pressure of 200 kPa and constant volume is heated from 30° ° C to 80° ° C, the pressure will be:

Question

A) 172kPa

B) 75 kPa

C) 233 kPa

D) 101.3 kPa

E) 533 kPa

Answer 1

We apply Gay-Lussac’s Law here. We convert temperature to the unit Kelvin.

P₁/T₁ = P₂/T₂

P₁ = 200kPa

T₁ = 30 + 273.15 = 303.15K

T₂ = 80 + 273.15 = 353.15K

P₂ = P₁ x T₂/T₁

200kPa x [353.15/303.15] = 233kPa = P₂

Answer 2

The ideal gas law: PV = nRT or PV/T = nR. If the gas is in a closed system then "nR" remains constant (because the number of molecules is the same). So at two different times, we can say:

P₁V₁/T₁ = nR = P₂V₂/T₂ and, in this case, since the volume stays constant, we can say:

P₁/T₁ = P₂/T₂

But to use this equation requires temperature to be in Kelvin. So T₁ = 30° + 273.15° = 303.15K and T₂ = 80° + 273.15° = 353.15

Solving P₁/T₁ = P₂/T₂ for P₂ gives: P₂ = P₁T₂/T₁ = 200(353.15)/303.15 = 233 kPa

Answer 3

The volume is constant so pressure is directly proportional to temperature (K)

Temperature in K changes from 303 to 353 (add 273 to celsius)

So the new pressure will be 200 x (353 / 303) = 233 kPa