A volume of gas V at temperature T1 and pressure P is enclosed in a sphere. If is connected to another sphere of volume V/2 by a tube and stop cork. This second sphere was initially evacuated and the stop cork is closed. If the stop cork is now opened and the temperature of the gas in the second sphere becomes T2 while the first sphere is maintained at the temperature of T1, show that the final pressure P within the apparatus is P'= 2PT2/2T2+T1
To understand and derive the relationship between the final pressure (P') within the apparatus and the temperatures (T1 and T2), we can make use of the ideal gas law, which states:
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 of the gas
Let's go step by step to derive the relationship:
Step 1: Consider the initial state of the system before the stop cork is opened.
- In the first sphere, we have an initial volume V, temperature T1, and pressure P.
- In the second sphere, we have an initial volume V/2, which is initially evacuated (no gas is present). The temperature is not relevant at this point as there is no gas.
Therefore, we can write the initial state equation for the first sphere as:
P1V = n1RT1 ---------- (Equation 1)
Step 2: When the stop cork is opened and the gases mix, we assume that the overall volume becomes V + V/2 = 3V/2 (cork opened).
Step 3: Now, let's consider the final state of the system after the gases mix and attain thermal equilibrium.
- The first sphere remains at temperature T1, so the final equation for the first sphere remains the same:
P1V = n1RT1 ---------- (Equation 1)
- In the second sphere, the gas attained temperature T2, so we can write the final equation for the second sphere as:
P2 * (V/2) = n2 * R * T2 ---------- (Equation 2)
Step 4: Combine the equations to find the final pressure (P').
Since the two spheres are connected, the total number of moles of gas (n) remains constant. So, we can write n1 = n2 = n.
Combining Equation 1 and Equation 2, we have:
P1V + P2 * (V/2) = n * R * T1 + n * R * T2
Simplifying further:
P1V + (P2 * V)/2 = n * R * (T1 + T2)
P1V + P2 * V/2 = 3nRT / 2
Rearranging the equation, we get:
(P1V + P2 * V/2) * 2 = 3P * V * T
2P1V + P2V = 3PVT
2P1 + P2 = 3P
Finally, substitute P2 = (P' * V)/2, as the final pressure within the apparatus is P'. We get:
2P1 + (P' * V)/2 = 3P
Multiplying through by 2 to get rid of the fraction:
4P1 + P'V = 6P
Now, substitute V = 3V/2, which is the combined volume:
4P1 + (P' * 3V/2) = 6P
Simplifying further:
4P1 + 3P'V = 6P
Since V = nRT / P, we can substitute V in terms of n, R, and T:
4P1 + 3P'(nRT / P) = 6P
Cancelling terms and rearranging:
4P1 + 3P'nRT / P = 6P
4P1P + 3P'nRT = 6P²
Dividing through by P:
4P1 + 3P'nRT = 6P
Rearranging gives:
3P'nRT = 6P - 4P1
3P'nRT = 2(3P - 2P1)
Finally, divide both sides by 2(nRT):
3P'nRT / 2(nRT) = (3P - 2P1) / 2(nRT)
Simplifying gives:
P' = (2P(3T) - 2P1(2T)) / (3T + 2T)
P' = (6PT - 4P1T) / 5T
P' = 2PT / (2T + T1)
Hence, we have derived the relationship between the final pressure (P') within the apparatus and the temperatures (T1 and T2):
P' = 2PT2 / (2T2 + T1)