A sample of gas at 1405 torr is cooled from 150.0 °C to 100.0 °C. Assuming the volume is constant, what is the pressure (in atm) of the gas at 100.0 °C?

Question

It would be very helpful if I could see this done step by step, I am getting a bit confused. Thank you!

Answer 1

For constant volume, the Ideal Gas Law simplifies into:

which is also known as Gay-Lussac's Law.

One stumbling block people sometimes have is that temperature in this equation must be in absolute temp (K) not in °C. To convert, add 273.15 to °C which makes T₁ = 423.15 K and T₂ = 373.15 K. Additionally, one pressure is given in torr but the final pressure is asked to be given in atmospheres (atm) so the easiest way to deal with that is just convert P₁ to units of atm. There are 760 torr in 1 atm so to convert 1405 torr, divide it by 760 to get 1.8487 atm

Multiplying both sides of Gay-Lussac's Law by T₂ we get:

P₂ = P₁T₂/T₁

Then plugging in our numbers gives us:

P₂ = (1.8487)(373.15)/(423.15) = 1.630 atm