`1 ` 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 J
B. 769 J
C. 1374 J
D. 969 J

Explanation

To solve the problem of finding the change in internal energy during the reversible adiabatic expansion of 1 mole of an ideal gas at STP, we can follow these steps:

### Step 1: Understand the Initial Conditions
At Standard Temperature and Pressure (STP), the initial temperature \( T_1 \) is given as:
\[ T_1 = 273.15 \, \text{K} \]
The initial volume \( V_1 \) is \( V \) (let's denote it as \( V_1 = V \)).

### Step 2: Determine the Final Volume
The problem states that the gas expands to double its volume:
\[ V_2 = 2V_1 = 2V \]

### Step 3: Use the Adiabatic Condition
For a reversible adiabatic process, the relationship between pressure and volume is given by:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
where \( \gamma = 1.4 \).

### Step 4: Relate Temperature and Volume
From the ideal gas law, we know:
\[ PV = nRT \]
For 1 mole of gas, we can express pressure in terms of temperature and volume:
\[ P = \frac{nRT}{V} \]
Substituting this into the adiabatic condition gives us:
\[ \frac{nRT_1}{V_1} V_1^\gamma = \frac{nRT_2}{V_2} V_2^\gamma \]
This simplifies to:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

### Step 5: Substitute Known 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 6: Solve for Final Temperature \( T_2 \)
Substituting \( \gamma = 1.4 \):
\[ T_1 = T_2 \cdot 2^{0.4} \]
Now substituting \( T_1 = 273.15 \, \text{K} \):
\[ 273.15 = T_2 \cdot 2^{0.4} \]
Solving for \( T_2 \):
\[ T_2 = \frac{273.15}{2^{0.4}} \approx 207 \, \text{K} \]

### Step 7: Calculate Change in Internal Energy
The change in internal energy for an ideal gas during an adiabatic process is given by:
\[ \Delta U = n C_V (T_1 - T_2) \]
Where \( C_V = \frac{R}{\gamma - 1} \). For 1 mole of gas:
\[ C_V = \frac{R}{0.4} \quad \text{(since } \gamma - 1 = 0.4\text{)} \]
Substituting \( R = 8.31 \, \text{J/(mol K)} \):
\[ C_V = \frac{8.31}{0.4} = 20.775 \, \text{J/K} \]
Now substituting \( T_1 \) and \( T_2 \):
\[ \Delta U = 1 \cdot 20.775 \cdot (273.15 - 207) \]
Calculating:
\[ \Delta U = 20.775 \cdot 66.15 \approx 1374 \, \text{J} \]

### Final Answer
The change in internal energy \( \Delta U \) is approximately:
\[ \Delta U \approx 1374 \, \text{J} \]

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