Gas is confined in a tank at a pressure of 11 ATM and a temperature of 25 degree Celsius. If 2/3 of the gas is withdrawn and the temperature is raised to 75 degree Celsius, what is the new pressure in the tank?
P V = n R T
11 = (n R/V) 298
so n R/V = 11/298
later
P V = (n/3) R T
3 P V = n R 348
so
n R/V = 3 P/348
but n R / V is still 11/298
so
(11/298)(348/3) = P
4.2804
To find the new pressure in the tank, we can use the combined gas law equation:
(P₁V₁) / (T₁) = (P₂V₂) / (T₂)
where:
P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure (to be calculated)
V₂ = final volume (unknown but since 2/3 of the gas is withdrawn, it will be 1/3 of the initial volume)
T₂ = final temperature
Given:
P₁ = 11 ATM
T₁ = 25 °C = 298 K
T₂ = 75 °C = 348 K
Since 2/3 of the gas is withdrawn, the final volume V₂ is 1/3 of the initial volume V₁.
Now, we need to rearrange the equation to solve for P₂:
P₂ = (P₁V₁T₂) / (V₂T₁)
Calculating:
P₂ = (11 ATM * V₁ * 348 K) / (1/3 * V₁ * 298 K)
Simplifying:
P₂ = (11 ATM * 348 K) / (1/3 * 298 K)
P₂ = (11 ATM * 348 K) / (0.3333 * 298 K)
P₂ ≈ 40.30 ATM
Therefore, the new pressure in the tank, when 2/3 of the gas is withdrawn and the temperature is raised to 75 °C, is approximately 40.30 ATM.
To find the new pressure in the tank, we can use the combined gas law, which relates the initial and final states of a gas sample.
The combined gas law can be written as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
- P1 and P2 are the initial and final pressures, respectively.
- V1 and V2 are the initial and final volumes, respectively.
- T1 and T2 are the initial and final temperatures, respectively.
Now, let's break down the given information:
Initial Pressure (P1) = 11 ATM
Initial Temperature (T1) = 25 degrees Celsius
Final Temperature (T2) = 75 degrees Celsius
Withdrawn Fraction of Gas = 2/3
Since the volume does not change in this problem, we can assume that V1 and V2 are equal. Therefore, we can simplify the equation as:
P1 / T1 = P2 / T2
Now, let's substitute the given values:
P1 / (25 + 273.15) = P2 / (75 + 273.15)
Converting the temperatures from degrees Celsius to Kelvin by adding 273.15:
11 ATM / (298.15 K) = P2 / (348.15 K)
Simplifying further, we can cross multiply:
11 ATM * (348.15 K) = P2 * (298.15 K)
Solving for P2:
P2 = (11 ATM * 348.15 K) / (298.15 K)
P2 ≈ 12.83 ATM
Therefore, the new pressure in the tank, when 2/3 of the gas is withdrawn and the temperature is raised to 75 degrees Celsius, is approximately 12.83 ATM.