Two moles of an ideal gas have an initial kelvin temperature `T_i` and absolute pressure `P_i` the gas undergoes a reversible isothermal compression from an initial volume `V_i` to a final volume 0.5 `V_i`
How much heat is exchanged with the environment? If heat is exchanged, is it absorbed or released?

Options

A. `p_i V_i `(In a) , Released
B. `P_i V_i `(In 2) ,absorbed
C. `(0.5 ) P_i v_i ` released
D. `( 0.5 ) P_i V_i` absorbed

Answer

To solve the problem of how much heat is exchanged with the environment during the isothermal compression of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**:
- The gas undergoes a reversible isothermal compression. This means the temperature remains constant throughout the process.

2. **Identify Given Values**:
- Number of moles of gas, \( n = 2 \) moles.
- Initial temperature, \( T_i \) (in Kelvin).
- Initial volume, \( V_i \).
- Final volume, \( V_f = 0.5 V_i \).

3. **Work Done in Isothermal Process**:
- The work done \( W \) during an isothermal process can be calculated using the formula:
\[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]
- Here, \( V_f = 0.5 V_i \), so:
\[ W = nRT \ln\left(\frac{0.5 V_i}{V_i}\right) = nRT \ln(0.5) = nRT \ln\left(\frac{1}{2}\right) = -nRT \ln(2) \]

4. **First Law of Thermodynamics**:
- According to the first law of thermodynamics:
\[ \Delta Q = \Delta U + W \]
- For an isothermal process, the change in internal energy \( \Delta U = 0 \) (since temperature is constant).
- Therefore:
\[ \Delta Q = W \]

5. **Calculate Heat Exchanged**:
- Substitute the expression for work into the equation for heat exchanged:
\[ \Delta Q = -nRT \ln(2) \]

6. **Interpret the Result**:
- The negative sign indicates that heat is released from the system to the environment during the compression process.

### Final Answer:
The amount of heat exchanged with the environment is:
\[ \Delta Q = -2RT_i \ln(2) \]
This heat is released (since \( \Delta Q < 0 \)).

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