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

Answer:

The volume of the gas is 11.2 L.

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:

$PV = nRT$

$V_{2} = \frac{nRT_{2}}{P_{2}}$    (1)

Where R is the gas constant

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

$n = \frac{P_{1}V_{1}}{RT_{1}}$    (2)

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

$V_{2} = \frac{P_{1}V_{1}}{RT_{1}}*\frac{RT_{2}}{P_{2}} = \frac{1 atm*700.0 L*32.0 ^{\circ}}{100.0 ^{\circ}*20.0 atm} = 11.2 L$

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

I hope it helps you!