`1 g` mole of an ideal gas at STP is subjected to a reversible adiabatic expansion to double its volume. Find the change in internal energy `( gamma = 1.4)`

Options

A. `1169.5 J`
B. `769.5 J`
C. `1369.5 J`
D. `969.5 J`

Explanation

To solve the problem of finding the change in internal energy of 1 g mole of an ideal gas undergoing a reversible adiabatic expansion to double its volume, we can follow these steps:

### Step 1: Understand the given information
- We have 1 g mole of an ideal gas.
- The process is a reversible adiabatic expansion.
- The initial conditions are at Standard Temperature and Pressure (STP), which means:
- \( T_1 = 273 \, K \)
- \( P_1 = 1 \, atm \)
- \( V_1 = 22.4 \, L \) (for 1 mole of gas at STP)
- The final volume \( V_2 = 2V_1 \).

### Step 2: Apply the adiabatic condition
For an adiabatic process, we can use the relation:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Where \( \gamma = 1.4 \).

### Step 3: Substitute the values
Substituting \( V_2 = 2V_1 \) into the equation:
\[ T_1 V_1^{\gamma - 1} = T_2 (2V_1)^{\gamma - 1} \]
This simplifies to:
\[ T_1 = T_2 \cdot 2^{\gamma - 1} \]

### Step 4: Solve for \( T_2 \)
Rearranging gives:
\[ T_2 = \frac{T_1}{2^{\gamma - 1}} \]
Substituting \( T_1 = 273 \, K \) and \( \gamma = 1.4 \):
\[ T_2 = \frac{273}{2^{0.4}} \]

### Step 5: Calculate \( 2^{0.4} \)
Calculating \( 2^{0.4} \):
\[ 2^{0.4} \approx 1.319 \]
Thus:
\[ T_2 \approx \frac{273}{1.319} \approx 206.6 \, K \]

### Step 6: Calculate the change in internal energy
The change in internal energy for an ideal gas during an adiabatic process is given by:
\[ \Delta U = nC_v(T_2 - T_1) \]
Where \( C_v = \frac{R}{\gamma - 1} \).

For 1 mole of gas:
\[ C_v = \frac{8.31}{1.4 - 1} = \frac{8.31}{0.4} = 20.775 \, J/K \]

### Step 7: Substitute values into the change in internal energy equation
Now substituting \( n = 1 \), \( C_v \), \( T_1 \), and \( T_2 \):
\[ \Delta U = 1 \times 20.775 \times (206.6 - 273) \]
Calculating \( (206.6 - 273) = -66.4 \):
\[ \Delta U = 20.775 \times (-66.4) \approx -1379.5 \, J \]

### Final Answer
The change in internal energy \( \Delta U \) is approximately:
\[ \Delta U \approx -1379.5 \, J \]