In reversible isothermal expansion of an ideal gas :

Options

A. w = 0
B. `U_(1) = U_(2)`
C. `H_(1) = H_(2)`
D. `q=nRTln.(V_(2))/(V_(1))`

Explanation

To solve the problem of reversible isothermal expansion of an ideal gas, we can follow these steps:

### Step 1: Understand the System
In a reversible isothermal expansion, the temperature (T) of the gas remains constant throughout the process. For an ideal gas, the internal energy (U) depends only on temperature.

**Hint:** Remember that for an ideal gas, internal energy is a function of temperature only.

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
Where:
- \( \Delta U \) = change in internal energy
- \( Q \) = heat added to the system
- \( W \) = work done by the system

**Hint:** Identify the terms in the equation and remember that for isothermal processes in ideal gases, the change in internal energy (\( \Delta U \)) is zero.

### Step 3: Determine Change in Internal Energy
Since the process is isothermal, the change in internal energy (\( \Delta U \)) is zero:
\[ \Delta U = 0 \]

Thus, we can rewrite the first law as:
\[ 0 = Q - W \]
This implies:
\[ Q = W \]

**Hint:** Recognize that if \( \Delta U = 0 \), then heat added to the system is equal to the work done by the system.

### Step 4: Calculate Work Done (W)
For a reversible isothermal expansion of an ideal gas, the work done (W) can be expressed as:
\[ W = -\int_{V_1}^{V_2} P_{\text{ext}} dV \]
For an ideal gas, using the equation of state \( PV = nRT \), we can express the work done as:
\[ W = nRT \ln \frac{V_2}{V_1} \]

**Hint:** Use the ideal gas law to relate pressure, volume, and temperature.

### Step 5: Calculate Heat Added (Q)
Since we established that \( Q = W \), we substitute the expression for work:
\[ Q = nRT \ln \frac{V_2}{V_1} \]

**Hint:** This shows that the heat added to the system during isothermal expansion is equal to the work done by the gas.

### Step 6: Change in Enthalpy (ΔH)
For an ideal gas, the change in enthalpy (\( \Delta H \)) is given by:
\[ \Delta H = nC_p \Delta T \]
Since the temperature is constant in an isothermal process, \( \Delta T = 0 \), thus:
\[ \Delta H = 0 \]

This means that the change in enthalpy is also zero:
\[ H_2 - H_1 = 0 \]
or
\[ H_2 = H_1 \]

**Hint:** Remember that enthalpy change for an ideal gas at constant temperature is zero.

### Final Summary
- The heat added to the system during the reversible isothermal expansion is \( Q = nRT \ln \frac{V_2}{V_1} \).
- The work done by the system is \( W = nRT \ln \frac{V_2}{V_1} \).
- The change in internal energy \( \Delta U = 0 \).
- The change in enthalpy \( \Delta H = 0 \).