Question

A gas is contained in a thick-walled balloon. When the pressure changes from 2.95 atm to atm, the volume changes from 7.456 L to 4.782 L and the temperature changes from 379 K to 212 K

Answer 1

Let's assume that the gas in the balloon is an ideal gas.

We can use combined gas law,

PV/T = k (constant)

Where, P is the pressure of the gas, V is volume of the gas and T is the temperature of the gas in Kelvin.

For two situations, we can use that as,

P₁V₁/T₁= P₂V₂/T₂

P₁ = 2.95 atm

V₁ = 7.456 L

T₁ = 379 K

P₂ = ?

V₂ = 4.782 L

T₂ = 212 K

By applying the formula,

2.95 atm x 7.456 L / 379 K = P₂ x 4.782 L / 212 K

P₂ = (2.95 atm x 7.456 L x 212 K) / (379 K x 4.782 L)

P₂ = 2.57 atm

Hence, the answer is 2.57 atm.