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:
[tex] PV = nRT [/tex]
[tex]V_{2} = \frac{nRT_{2}}{P_{2}}[/tex] (1)
Where R is the gas constant
With the initials conditions we can find the number of moles:
[tex] n = \frac{P_{1}V_{1}}{RT_{1}} [/tex] (2)
By entering equation (2) into (1) we have:
[tex] 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 [/tex]
Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.
I hope it helps you!