A large flexible balloon contains 1.5moles of a gas in a volume of 27liters. If 1.1moles of the gas are removed and the pressure and temperature do not change, what will be the new volume of the gas?

Question

Round to 2 significant figures.

Answer 1

The ideal gas law is PV=nRT. Using simple algebra, we can rearrange this to be P/T = nR/V. If Pressure and temperature stay constant then we can say n₁R/V₁ = P/T = n₂R/V₂ or

n₁R/V₁ = n₂R/V₂

Since R is a constant, we can divide both sides by R and get:

n₁/V₁ = n₂/V₂

Again, using simple algebra, we can rearrange this to be:

V₂ = V₁n₂/n₁

So, since n₂ = 1.5 - 1.1 = 0.4 we can plug in the numbers to get:

V₂ = (27)(0.4)/(1.5) = 7.2 L

Answer 2

You can solve this problem using the ideal gas law:

PV = nRT where

P is the pressure of the gas

V is the volume of the gas

n is the moles of the gas

R is a constant

and T is the temperature of the gas

Initially, you 1.5 mol of gas and it occupies 27 L. Therefore:

P*(27 L) = (1.5 mol)*R*T

If you remove 1.1 mol of gas, you're left with 0.4 mol of gas. Therefore:

P*(x L) = (0.4 mol) *R*T

Since P, R, and T remain constant, you can set up fractions and cross multiply and divide:

27 L x L

_______ = _______

1.5 mol 0.4 mol

=> (27 L)*(0.4 mol)/(1.5 mol) = 7.2 L