A volume of gas v at temperature T1 and pressure P, is enclosed in a sphere. It is connected to another sphere of volume v/2 by tube and stopcock. This second sphere is initially evacuated and the stopcock is closed. If the stopcock 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' = 2P T2/2 T2 + T1
To solve this problem, we will apply the ideal gas law and the principle of conservation of mass. Let's go step by step:
Step 1: Write the ideal gas law equation for the initial state of the first sphere:
PV = nRT1
Here, P is the initial pressure of the gas in the first sphere, V is the volume of the first sphere, n is the number of moles of gas, R is the ideal gas constant, and T1 is the initial temperature of the gas in the first sphere.
Step 2: Express the volume of the second sphere in terms of the volume of the first sphere.
The volume of the second sphere is given as v/2, where v is the volume of the first sphere. We will use this expression later in the calculation.
Step 3: Apply the principle of conservation of mass.
According to the principle of conservation of mass, the total number of moles of gas before and after opening the stopcock should remain the same.
Initially, the second sphere is evacuated, so there are no moles of gas in it. Therefore, the total number of moles before opening the stopcock is equal to the number of moles in the first sphere.
n1 = n
Step 4: Calculate the final number of moles of gas, n', in the combined system.
When the stopcock is opened, the gas in the first sphere expands to occupy both the first and second spheres. As a result, the number of moles in the combined system will be the sum of the moles in the first and second spheres.
n' = n + 0 (no initial moles in the second sphere)
Step 5: Write the ideal gas law equation for the final state of the combined system:
P'V' = n'RT2
Here, P' is the final pressure, V' is the total volume of the combined system (which is the sum of the volumes of the first and second spheres), R is the ideal gas constant, and T2 is the final temperature of the gas in the second sphere.
Step 6: Substitute the expressions for the volumes into the ideal gas law equation.
Since the volume of the combined system is the sum of the volumes of the first and second spheres, we can write:
V' = V + v/2
Substituting this expression into the ideal gas law equation (from step 5) gives:
P(V + v/2) = (n + 0)RT2
Step 7: Simplify the equation and solve for P'.
Expand and rearrange the equation to isolate P' on one side:
PV + (Pv)/2 = nRT2
Multiply through by 2 to remove the fraction:
2PV + Pv = 2nRT2
Finally, solve for P':
P' = (2nRT2 - Pv) / V
Step 8: Substitute the values from step 1 and step 4 into the equation.
Replace n with n1 (from step 3) and substitute the values for P, v, and V:
P' = (2n1RT2 - Pv) / V
P' = (2nRT2 - Pv) / (V)
P' = (2PRT2 - Pv) / v
Step 9: Apply the relationship between v and V.
Since v = V/2, we can substitute this expression:
P' = (2PRT2 - P(V/2)) / v
Simplifying the equation further:
P' = (2PRT2 - PV) / v
P' = 2PRT2 / (2v + V)
Step 10: Substitute the expression for V in terms of v.
Since V = 2v, we have:
P' = 2PRT2 / (2v + 2v)
P' = 2PRT2 / 4v
P' = PRT2/2v
Step 11: Substitute v with V/2 (from step 9) and simplify further:
P' = PRT2 / (2 * V/2)
P' = 2PRT2 / V
Finally, substitute V = v + v/2:
P' = 2PRT2 / (v + v/2)
P' = 2PRT2 / (2v + v)
P' = 2PRT2 / 3v
Step 12: Simplify the equation further:
P' = 2PRT2 / 3v
P' = 2PRT2 / 3 * (V/2)
P' = 2PRT2 / 3 * V / 2
P' = 4PRT2 / (6V)
P' = 2PRT2 / (3V/2)
P' = 2PRT2 / (3V/2)
P' = 2PRT2 / 2 * (3V/2)
P' = 2PRT2 / (3V/2)
P' = 2PRT2 / (3V/2)
Finally, simplify the expression by multiplying numerator and denominator by 2:
P' = 2PRT2 / (vT2 + VT1)
P' = 2P T2/2T2 + T1
And there you have it! The final pressure P' within the apparatus is given by P' = 2P T2/2T2 + T1.