Question

700.0 liters of a gas are prepared at 760.0 mmHg and 100.0 °C. The gas is placed into a tank under high pressure. When the tank cools to 32.0 °C, the pressure of the gas is 20.0 atm. What is the volume of the gas?

Answer 1

The volume of the gas is 11.2 L.

Step-by-step explanation:

Initially, we have:

V₁ = 700.0 L

P₁ = 760.0 mmHg = 1 atm

T₁ = 100.0 °C

When the gas is in the thank we have:

V₂ =?

P₂ = 20.0 atm

T₂ = 32.0 °C

Now, we can find the volume of the gas in the thank by using the Ideal Gas Law:

`PV = nRT`

`V_(2) = (nRT_(2))/(P_(2))` (1)

Where R is the gas constant

With the initials conditions we can find the number of moles:

`n = (P_(1)V_(1))/(RT_(1))` (2)

By entering equation (2) into (1) we have:

`V_(2) = (P_(1)V_(1))/(RT_(1))*(RT_(2))/(P_(2)) = (1 atm*700.0 L*32.0 ^(\circ))/(100.0 ^(\circ)*20.0 atm) = 11.2 L`

Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.

I hope it helps you!