A gas is originally kept in a 15.00 L container at 40° C and 1.67 atm. If the temperature is increased to 70° C and the volume decreased to 7.49 L, what is the new pressure?

Answer 1

Given:

V1 = 15.00 L

T1 = 40 + 273 = 313K

P1 = 1.67 atm

T2 = 70 + 273 = 343K

V2 = 7.49 L

Asked for:

P2

Combined gas law: P1V1/T1 = P2V2/T2

Solving for P2 we have...

P1V1T2/T1V2 = P2

P2 = (1.67)(15)(343) / (313)(7.49)

P2 = 3.67 atm

Answer 2

Find an answer here.

https://youtu.be/7HLHSUS6bGM

Answer 3

You want to use the combined gas law formula for this, which says that:

P₁V₁ = P₂V₂

T₁ T₂

To use the formula, temperature has to be in degrees Kelvin so add 273° to each temperature.

Do you think you'd be able to solve for the new pressure?