Use the combined gas law or ideal gas equation please

Question

There is a gas at a pressure of 12.0 atm, a volume of 23.0 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas? - Show Work

Answer 1

Here you need to use the ideal gas equation: PxV = nxRxT 

You assume that the gas is ideal which means that the ideal equation can be applied at any state. Before heating, state 1: P₁xV₁ = nxRxT₁.After heating, state 2: P₂xV₂ = nxRxT₂ 

P₁xV₁/P₂V₂ = (nxRxT₁/nxRxT₂) n, and R will be deleted, and then:

P₁xV₁/P₂V₂ = T₁/T₂

 

V₂ = P₁xV₁xT₂/P₂xT₁= (12.0x23.0x300)/(14.0x200) = 29.6 L 

 

Answer 2

If the amount of the gas is constant, then

 

P₁V₁/T₁ = P₂V₂/T₂

 

P₁ = 12.0 atm

 

V₁ = 23.0 L

 

T₁ = 200 K

 

P₂ = 14 atm

 

T₂ = 300 K

 

V₂ = (T₂/T₁) (P₁V₁/P₂) = (300/200)(12.0*23.0/14) L = 29.6 L