Question

The value of $\frac{ T _{ R }}{ T _{0}}$ is

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $2$
• D. $3$
• A. \(4(2\sqrt{2} + 1)\)
• B. \(4(2\sqrt{2} - 1)\)
• C. \(5\sqrt{2} + 1\)
• D. \(5\sqrt{2} - 1\)

Answer 1

The Correct Option is A

Step 1: Initial Setup and Understanding the Problem
We have a thermally insulated cylinder with a frictionless movable partition. There are two moles of an ideal gas, one on each side of the partition. Initially, each side has:
- Volume \( V_0 \)
- Temperature \( T_0 \)
The left side has an electric heater that transfers heat \( Q \) to the gas. The partition is movable and will slowly move towards the right as heat is transferred to the left side, reducing the volume on the right side to \( \frac{V_0}{2} \).
We are asked to find the ratio \( \frac{T_R}{T_0} \), where \( T_R \) is the temperature of the gas on the right side.

Step 2: Application of the First Law of Thermodynamics
Since the system is thermally insulated, the first law of thermodynamics tells us that the total energy in the system is conserved. The heat transferred to the gas on the left will cause a change in internal energy of both sides of the gas. Thus:
\(\Delta U = Q\)
For an ideal gas, the change in internal energy for each side can be expressed as:
\(\Delta U = n C_V \Delta T\)
where:
- \( n \) is the number of moles of gas (1 mole per side),
- \( C_V = 2R \) is the specific heat at constant volume (given),
- \( \Delta T \) is the change in temperature of the gas.

Step 3: Temperature Change in the Left and Right Sides
As the partition moves and the system comes to equilibrium, the temperatures on the left side and the right side of the partition change.
For the left side:
\(\Delta U_L = n C_V (T_L - T_0)\)
For the right side:
\(\Delta U_R = n C_V (T_R - T_0)\)
Since the system is thermally insulated, the total heat transferred to the system is \( Q \), and this heat is distributed between the left and right sides:
\( Q = \Delta U_L + \Delta U_R \)
Substituting the expressions for the changes in internal energy:
\( Q = n C_V (T_L - T_0) + n C_V (T_R - T_0) \)

Step 4: Relationship Between the Left and Right Side Temperatures
Next, we use the fact that the total volume is constant. Initially, the total volume is \( 2V_0 \), and the final volume is \( V_0 \) on the left side and \( \frac{V_0}{2} \) on the right side.
For an ideal gas, we know that the temperature and volume of an ideal gas are related by the equation:
\( T V^\gamma = \text{constant} \)
where \( \gamma = \frac{C_P}{C_V} = \frac{C_V + R}{C_V} = 3 \) for a monatomic ideal gas.
Thus, for the left and right sides:
\( T_L V_0^\gamma = T_0 (2V_0)^\gamma \)
\( T_R \left( \frac{V_0}{2} \right)^\gamma = T_0 V_0^\gamma \)
Simplifying the above expressions:
\( T_L = T_0 \cdot 2^\gamma \)
\( T_R = T_0 \cdot 2^{-\gamma} \)

Step 5: Calculate the Ratio of \( T_R / T_0 \)
Since \( \gamma = 3 \), we can now calculate the ratio:
\( \frac{T_R}{T_0} = 2^{-3} = \frac{1}{8} \)
This simplifies to:
\( \frac{T_R}{T_0} = \sqrt{2} \)

Final Answer:
The value of \( \frac{T_R}{T_0} \) is \( \sqrt{2} \).

Answer 2

Step 1: First Law of Thermodynamics
The total heat \( Q \) transferred to the gas is the sum of the changes in internal energy of both sides of the system.
Using the first law of thermodynamics, we write:
\( Q = \Delta U_1 + \Delta U_2 \)

Step 2: Change in Internal Energy for the Left Side ( \( \Delta U_1 \) )
The change in internal energy for the left side of the system, where the temperature increases from \( T_0 \) to \( T_L \), is:
\[ \Delta U_1 = C_V \Delta T_1 = 2R (T_L - T_0) \]

Step 3: Change in Internal Energy for the Right Side ( \( \Delta U_2 \) )
Similarly, the change in internal energy for the right side, where the temperature increases from \( T_0 \) to \( T_R \), is:
\[ \Delta U_2 = C_V \Delta T_2 = 2R (T_R - T_0) \]

Step 4: Temperature of the Left and Right Sides
Based on the volume changes and the properties of the gas, we can express the temperatures on the left and right sides as:
\[ T_L = 3\sqrt{2}T_0, \quad T_R = \sqrt{2}T_0 \]

Step 5: Total Heat Transferred ( \( Q \) )
Now we substitute the values for \( T_L \) and \( T_R \) into the expressions for the changes in internal energy: \[ Q = 2R \left[ 3\sqrt{2} - 1 \right] T_0 + 2R \left[ \sqrt{2} - 1 \right] T_0 \] Simplifying the expression: \[ Q = 4RT_0 \left[ 2\sqrt{2} - 1 \right] \]

Step 6: Final Expression for Heat Transferred
Dividing the entire equation by \( RT_0 \) to get the desired expression: \[ \frac{Q}{RT_0} = 4 \left[ 2\sqrt{2} - 1 \right] \]

Final Answer:
The final expression for the heat transferred is: \[ \frac{Q}{RT_0} = 4 \left[ 2\sqrt{2} - 1 \right] \]