How much air would have to be forced into the tank to increase the pressure, assuming no change in temperature?

Question

An air storage tank whose volume is 110 liters contains 2kg of air at a pressure of 15 atmospheres. How much air would have to be forced into the tank to increase the pressure to 18 atmospheres assuming no change in temperature?

Answer 1

Hello, Zehra,

We need to use the ideal gas law PV = nRT. We have two conditions, which I'll note with the subscripts 1 and 2, for the initial and final conditions.

The values for each are:

P₁ = 15atm

V₁ = 110L

n₁ = 2kg

T₁ = constant

P₂ = 18atm

V₂ = 110L

n₂ = X

T₂ = constant

The values for each are:

P₁ =

P₁V₁ = n₁RT₁ and P₂V₂ = n₂RT₂

Note that neither the volume nor the temperature change, so V₁ = V₂ and T₁ = T₂.

We can divide one equation by the other, and cancel the values that do not change:

(P₁V₁)/(P₂V₂) = (n₁RT₁)/(n₂RT₂)

Cancel the volumes and temperatures, since they do not change.

(P₁)/(P₂) = (n₁)/(n₂)

I'll invert this to get the ratio n₂/n₁

n₂/n₁ = P₂/P₁

n₂/n₁ = 18atm/15atm

n₂/n₁ = 1.2

2kg * 1.2 = 2.4kg air needed to increase pressure to 18 atm.

Note that n₁ and n₂ should both be in units of moles for the ideal gas law, but we are looking at the ratio of n₂ to n₁ and all gases have a value of 22.4 L/mole at STP. So the ratio of n₁/n₂ should also be the ratio of those gases in mass, since we aren't changing the composition ("air").

I hope this helps,

Bob

Answer 2

Since the volume of the tank is constant as is the temperature, we have the following relationship:

P1/n1 = P2/n2

P1 = initial pressure = 15 atm

n1 = initial moles of air = 2 kg x 1000 g/kg x 1 mol/29 g = 69 moles (assuming a molar mass of 29 g/mol)

P2 = 18 atm

n2 = final moles = ?

Solving for n2 we have...

P2(n1) / P1 = n2

n2 = (18 atm)(69 mol) / 15 atm

n2 = 83 moles

Since the tank already contains 69 moles, the amount needed to be added is 83 mol - 69 mol = 14 moles

14 moles x 29 g /mol = 406 g of air need to be forced into the tank (not corrected for sig. figs.)

Corrected for sig. figs. you have 400 g = 0.4 kg

Answer 3

pv = m/µRT

p₁V = m₁RT

From here p/p₁ = m/m₁

m₁ = mp₁/p = 2kg·18atm/15atm = 2.4kg

Δm = 2.4kg - 2kg = 0.4 kg