You have a tank of argon gas at 19.80 atm pressure at 19°C. The volume of argon in the tank is 50.0 L. What would be the volume (in liters) of this gas if you allowed it to expand to the pressure of the surrounding air (0.924 atm)? Assume the temperauter remains constant.
P1V1 = P2V2
Substitute and solve for V2.
To find the volume of the argon gas when it expands to the pressure of the surrounding air, you can use the combined gas law. The combined gas law formula is given as:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Where:
P₁ = Initial pressure of the gas
V₁ = Initial volume of the gas
T₁ = Initial temperature of the gas (constant in this case)
P₂ = Final pressure of the gas
V₂ = Final volume of the gas (what we want to find)
T₂ = Final temperature of the gas (constant in this case)
Let's plug in the known values into the formula:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Initially, we have:
P₁ = 19.80 atm
V₁ = 50.0 L
T₁ = 19°C + 273.15 = 292.15 K (converting from Celsius to Kelvin)
The final condition is:
P₂ = 0.924 atm
Since the temperature remains constant, T₂ = T₁ = 292.15 K
Now, let's solve for V₂:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
(19.80 atm * 50.0 L) / 292.15 K = (0.924 atm * V₂) / 292.15 K
(19.80 atm * 50.0 L) = (0.924 atm * V₂)
V₂ = (19.80 atm * 50.0 L) / 0.924 atm
V₂ ≈ 1069.57 L
Therefore, the volume of the argon gas when it expands to the pressure of the surrounding air would be approximately 1069.57 liters.