15. Two 5 dm3 flasks are connected by by a narrow tube of negligible volume. Initially the two flasks are both at 27 oC and contain a total of 2 moles of an ideal gas. One flask is heated to a uniform temperature of 127 oC while the other is kept at 27 oC. Assuming their volume does not alter, calculate the number of moles of gas in each flask of the gas and the final pressure.
Poorly explained
To solve this problem, we need to apply the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
Given that the volume is constant and the two flasks are connected by a narrow tube, the pressure in both flasks will be the same after equilibrium is reached.
Let's solve the problem step-by-step:
Step 1: Find the number of moles in each flask at the initial temperature.
The number of moles of gas is given as 2 moles.
Step 2: Calculate the final pressure when the system reaches equilibrium.
To find the final pressure, we can use the equation:
(P1 * V1) + (P2 * V2) = (n1 * R * T1) + (n2 * R * T2)
Since the two flasks are connected and the volume is constant, we can rewrite the equation as:
(P1 + P2) * V = n1 * R * T1 + n2 * R * T2
Substituting the given values:
P1 = ?
P2 = ?
V = 5 dm^3 (the volume of each flask)
n1 = 2 moles (initial number of moles)
n2 = 0 moles (initial number of moles in the second flask)
R = ideal gas constant (8.314 J/(mol*K))
T1 = 127 °C (temperature of the heated flask)
T2 = 27 °C (temperature of the other flask)
Converting the temperatures to Kelvin:
T1 = 127 + 273 = 400 K
T2 = 27 + 273 = 300 K
Plugging in the values into the equation, we can solve for the final pressure:
(P1 + P2) * 5 = (2 * 8.314 * 300) + (0 * 8.314 * 400)
5P1 + 5P2 = 4988.4
Step 3: Calculate the number of moles of gas in each flask at equilibrium.
Since the number of moles in both flasks will be the same at equilibrium, we can write:
n1 = n2
Let's assume both flasks have n moles of gas.
From the ideal gas law equation, we know that:
P1 * V = n * R * T1
P2 * V = n * R * T2
Substituting the given values:
P1 = ?
P2 = ?
V = 5 dm^3 (the volume of each flask)
R = ideal gas constant (8.314 J/(mol*K))
T1 = 127 °C (temperature of the heated flask)
T2 = 27 °C (temperature of the other flask)
n = ?
Converting the temperatures to Kelvin:
T1 = 127 + 273 = 400 K
T2 = 27 + 273 = 300 K
Now we have two equations with two unknowns:
P1 * 5 = n * 8.314 * 400
P2 * 5 = n * 8.314 * 300
Dividing the second equation by the first equation:
(P2 * 5) / (P1 * 5) = (n * 8.314 * 300) / (n * 8.314 * 400)
P2 / P1 = 3/4
We know that P1 + P2 = 4988.4 / 5 = 997.68.
With these two equations:
P1 + P2 = 997.68
P2 / P1 = 3/4
We can solve for P1 and P2.
Multiplying the second equation by P1:
P2 = (3/4) * P1
Substituting P2 in the first equation:
P1 + (3/4) * P1 = 997.68
(7/4) * P1 = 997.68
P1 = (4/7) * 997.68
P1 ≈ 569.54
Substituting P1 in the second equation:
P2 = (3/4) * 569.54
P2 ≈ 427.15
Therefore, at equilibrium, both flasks will contain approximately:
n1 = n2 = n ≈ 1.14 moles.
The final pressure will be approximately:
P = P1 + P2 ≈ 569.54 + 427.15 ≈ 996.69
So, there will be approximately 1.14 moles of gas in each flask, and the final pressure will be approximately 996.69 Pascal.
To solve this problem, we can use the ideal gas equation:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
First, let's convert the temperatures from degrees Celsius to Kelvin:
Initial temperature = 27 oC = 27 + 273.15 = 300.15 K
Final temperature = 127 oC = 127 + 273.15 = 400.15 K
Now, let's analyze the two flasks separately:
Flask 1 (heated flask):
Since the volume of each flask is 5 dm3 and does not alter, the volume of gas in Flask 1 remains 5 dm3.
We know the initial temperature (300.15 K), initial pressure (unknown), and the final temperature (400.15 K). Let's calculate the number of moles of gas using the initial information.
Using the ideal gas equation:
PV = nRT
We can solve for n:
n = (PV) / (RT)
Since the volume (V) and the gas constant (R) are constant, we can simplify the equation to:
n = (P / T)
Substituting the known values:
n1 = (P1 / T1)
Flask 2 (kept at 27 oC):
The volume of gas in Flask 2 remains 5 dm3.
The initial temperature is 300.15 K, and the final temperature is still 300.15 K (since it is not heated). Let's calculate the number of moles for Flask 2:
n2 = (P2 / T2)
Since the volumes of the two flasks are connected and the total number of moles is given as 2 moles, we can write the equation:
n1 + n2 = 2
Now, let's solve these equations simultaneously to determine the number of moles in each flask:
(P1 / T1) + (P2 / T2) = 2
To find the final pressure, we'll need to isolate P1 (the pressure in Flask 1) and substitute the known values into the equation.
Once we have the values of n1, n2, and P1, we can calculate the final pressure by combining the two flask volumes and using the ideal gas equation:
P_total = (n1 + n2) * R * T_total / V_total
Substituting the known values, we can find the final pressure.
Two 5 dm3 flasks are connected by a narrow tube of negligible volume. Initially the two flasks are both at 27oC and contain a total of 2 moles of an ideal gas. One flask is heated to a uniform temperature of 127oC while the other is kept at 27oC. Assuming their volume does not alter, calculate the number of moles of gas in each flask of the gas and the final pressure.
Solution: As their volumes are kept constant, P then varies directly with T. A number of 1 mole (= 2/1) of gas is comprehensively distributed within each flask.
In the initial condition on ideal gas, PV = nRT or P x (2x5) = 2 x 0.082 x (273+27) => P equals 4.92 atm as an initial one.
Anyway, as the flask of our consideration is heated to the temperature of 127 degree Celcius, a number of total moles are kept constant at 2 moles, but leaking from the flask of our consideration to the other one, for example X mol, leaving the flask 1 - X mole, and the other one 1 + X mol.
Heated flask: P’V = nRT
=> P’x V = (1 - X) x R x (273+127)
Likewise, tempt-controlling flask: P’V = nRT
=> P’ x V = (1 + X) x R x (273+27)
Then, X = 0.143 mol. That is, heated flask will contain gas of 1 - 0.143 = 0.857 mol and the other one 1.143 mol.
Considering the heated flask, PV = nRT => P x 5 = 0.857 x 0.082 x (273+127) = 5.62 atm as a final pressure gauged. We can check for the correctness by considering the other flask; PV = nRT => P x 5 = 1.143 x 0.082 x (273+27) = 5.62 atm.