An ideal gas, \(\overline{C_V} = \frac{5}{2} R\), is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume.
If the initial temperature and pressure are \(298 \, \text{K}\) and \(5 \, \text{atm}\), respectively, then the final temperature is ______ \( \, \text{K} \) (nearest integer).
Given: \(\overline{C_V}\) is the molar heat capacity at constant volume.
If the initial temperature and pressure are \(298 \, \text{K}\) and \(5 \, \text{atm}\), respectively, then the final temperature is ______ \( \, \text{K} \) (nearest integer).
Given: \(\overline{C_V}\) is the molar heat capacity at constant volume.
Correct Answer: 274
Approach Solution - 1
For an adiabatic process:
\[\Delta U = q + w, \quad q = 0 \implies \Delta U = w\]
Step 1: Using the first law of thermodynamics:
\[n C_V \Delta T = -P_{\text{ext}} (V_2 - V_1)\]
Since $V_2 = 2V_1$, substitute and simplify:
\[nR \frac{T_2}{P_2} = 2nR \frac{T_1}{P_1}.\]
Step 2: Relation between $P_2$ and $T_2$:
\[P_2 = \frac{5T_2}{2 \times 298}.\]
Step 3: Using $C_V$:
\[\frac{5}{2} nR (T_2 - T_1) = -nR T_1 \left( \frac{P_2}{P_1} - 1 \right).\]
Step 4: Substitute and solve:
\[T_2 = \frac{5T_2}{2 \times 298}.\]
From the equation:
\[T_2 \approx 274.16 \, \text{K}.\]
Nearest integer:
\[T_2 \approx 274 \, \text{K}.\]
Approach Solution -2
This problem involves calculating the final temperature of an ideal gas after it undergoes an irreversible adiabatic expansion against a constant external pressure.
Concept Used:
The solution is based on the First Law of Thermodynamics and the properties of an ideal gas.
- First Law of Thermodynamics: The change in internal energy (\(\Delta U\)) of a system is the sum of the heat added to the system and the work done on the system. \[ \Delta U = q + w \]
- Adiabatic Process: An adiabatic process is one where no heat is exchanged with the surroundings, so \(q = 0\). For such a process, the First Law simplifies to: \[ \Delta U = w \]
- Internal Energy of an Ideal Gas: The change in internal energy of \(n\) moles of an ideal gas depends only on the change in temperature and is given by: \[ \Delta U = n\overline{C_V}(T_2 - T_1) \] where \(\overline{C_V}\) is the molar heat capacity at constant volume.
- Work Done in Irreversible Expansion: The work done on the gas during an expansion against a constant external pressure (\(P_{ext}\)) is: \[ w = -P_{ext} (V_2 - V_1) \]
- Ideal Gas Law: The relationship between pressure, volume, and temperature for an ideal gas is \(PV = nRT\).
Step-by-Step Solution:
Step 1: List the given information.
- Molar heat capacity at constant volume: \(\overline{C_V} = \frac{5}{2}R\)
- Initial pressure: \(P_1 = 5 \, \text{atm}\)
- Initial temperature: \(T_1 = 298 \, \text{K}\)
- Constant external pressure: \(P_{ext} = 1 \, \text{atm}\)
- Volume relation: \(V_2 = 2V_1\)
- The process is adiabatic, so \(q = 0\).
Step 2: Apply the First Law of Thermodynamics for an adiabatic process.
Since \(q=0\), we have \(\Delta U = w\). Substituting the expressions for \(\Delta U\) and \(w\):
\[ n\overline{C_V}(T_2 - T_1) = -P_{ext}(V_2 - V_1) \]
Step 3: Substitute the given volume relationship \(V_2 = 2V_1\) into the equation.
\[ n\overline{C_V}(T_2 - T_1) = -P_{ext}(2V_1 - V_1) \] \[ n\overline{C_V}(T_2 - T_1) = -P_{ext}V_1 \]
Step 4: Use the Ideal Gas Law to express the initial volume \(V_1\) in terms of the initial conditions.
\[ P_1 V_1 = nRT_1 \implies V_1 = \frac{nRT_1}{P_1} \]
Substitute this expression for \(V_1\) into the equation from Step 3.
\[ n\overline{C_V}(T_2 - T_1) = -P_{ext}\left(\frac{nRT_1}{P_1}\right) \]
Step 5: Substitute the given value for \(\overline{C_V} = \frac{5}{2}R\) and simplify the equation. The terms \(n\) and \(R\) cancel out.
\[ n\left(\frac{5}{2}R\right)(T_2 - T_1) = -P_{ext}\left(\frac{nRT_1}{P_1}\right) \] \[ \frac{5}{2}(T_2 - T_1) = -\frac{P_{ext}}{P_1}T_1 \]
Step 6: Substitute the given numerical values into the simplified equation and solve for the final temperature \(T_2\).
\[ \frac{5}{2}(T_2 - 298) = -\frac{1 \, \text{atm}}{5 \, \text{atm}}(298 \, \text{K}) \] \[ \frac{5}{2}(T_2 - 298) = -\frac{298}{5} \] \[ T_2 - 298 = -\frac{298}{5} \times \frac{2}{5} \] \[ T_2 - 298 = -\frac{596}{25} \] \[ T_2 - 298 = -23.84 \] \[ T_2 = 298 - 23.84 = 274.16 \, \text{K} \]
The problem asks for the nearest integer value.
The final temperature is 274 K.
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