Question

A sample of gas at a pressure of 1.36 atm and a temperature of 20.1^oC is heated to 45.6^oC. If the gas expanded to 2.52L from 2.15L, what is the pressure of the heated gas?

Answer 1

Ideal gas equation is PV = nRT

Where

T = Temperature ;P = pressure ;n = No . of moles ; R = gas constant ;V= Volume of the gas

Since the sample of the gas doesn't change so number of moles( mass) will be the same.And also the value of R will not chages.

So For the first case P₁V₁ = nRT₁ --------------------- (1)

    In the second case P₂V₂=nRT₂ ------------------- ( 2)

Equ(1) / Eqn(2) gives

Where P₁= initial pressure = 1.36 atm

          V₁ = initial volume = 2.15 L

         T₁ = initial temperature = 20.1 ^oC = 20.1 + 273 =293.1 K

         P₂= final pressure = ?

         V₂ = final volume = 2.52 L

         T₂ = final temperature = 45.6 ^oC = 45.6 + 273 = 318.6 K

Plug the values we get

                                

So the pressure of the heated gas is 1.26 atm

NOTE: in the given problem the volume is reversed since upon heating the gas expands so that it increases from 2.15L to 2.52 L.Correct that one.