Final answer:
The final pressure of the gas remaining in the tank after two-thirds of the gas is withdrawn and the temperature is raised to 74.0°C is approximately 3.827 atm.
Step-by-step explanation:
To determine the pressure of the gas remaining in the tank after two-thirds of the gas has been withdrawn and the temperature raised, we can use the ideal gas law in a before-and-after scenario. The ideal gas law is represented by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in kelvins.
Firstly, since two-thirds of the gas is removed, only one-third remains. Consequently, we can infer that the number of moles (n) in the tank is now one-third of the original. Secondly, we use the fact that both initial and final pressures are measured when the gas occupies the same volume and use absolute temperatures for calculations.
Initial conditions:
Pressure (P1) = 10.3 atm
Temperature (T1) = 23.5°C = 296.65 K
Final conditions:
Pressure (P2) = ?
Temperature (T2) = 74.0°C = 347.15 K
The number of moles has reduced to one-third, therefore, n2 = n1/3.
Since the volume and the ideal gas constant do not change, we can state that P1/T1 = P2/T2.
Substituting the values, we get:
(10.3 atm) / (296.65 K) = P2 / (347.15 K)
P2 = (10.3 atm) * (347.15 K) / (296.65 K) * (1/3)
P2 ≈ 3.827 atm
The final pressure of the gas remaining in the tank after the given changes is approximately 3.827 atm.