In an closed vessel 300 torr of argon gas is filled and the vessel heated to 925 degree celcius from room temperature. What will be the pressure at 925 degree celcius .
P2/P1=T2/T1
solve for P2
change temps to Kelvins
In your case, P1 = 300 torr = 0.395 atm,
T1 = 295 K (22 C) (room temperature)
T2 = 1198 K
Note the correct spelling of celsius.
To determine the pressure of argon gas inside a closed vessel at 925 degrees Celsius, we can use the Ideal Gas Law equation:
P₁V₁/T₁ = P₂V₂/T₂
where P₁ and P₂ are the initial and final pressures, V₁ and V₂ are the initial and final volumes, and T₁ and T₂ are the initial and final temperatures (in Kelvin).
In this case, the initial pressure (P₁) is given as 300 torr, the initial temperature (T₁) is the room temperature (which we'll assume is 25 degrees Celsius or 298 Kelvin), and the final temperature (T₂) is 925 degrees Celsius or 1198 Kelvin. We are asked to find the final pressure (P₂).
Since the volume (V₁ and V₂) is not changing in this scenario (as the vessel is closed), we can eliminate the volume from the equation:
P₁/T₁ = P₂/T₂
Now, let's substitute the given values into the equation:
(300 torr) / (298 K) = P₂ / (1198 K)
To solve for P₂, we can cross-multiply and solve for the unknown:
P₂ = (300 torr) * (1198 K) / (298 K)
P₂ ≈ 1209.40 torr
Therefore, the pressure of argon gas inside the closed vessel at 925 degrees Celsius is approximately 1209.40 torr.