One mole of an ideal monoatomic gas is taken at a temperature of `300 K`. Its volume is doubled keeping its pressure constant. Find the change in internal energy.

Explanation

To find the change in internal energy when one mole of an ideal monoatomic gas is taken at a temperature of 300 K and its volume is doubled while keeping the pressure constant, we can follow these steps:

### Step 1: Understand the internal energy formula
The internal energy (U) of an ideal gas is given by the formula:
\[ U = \frac{f}{2} N R T \]
where:
- \( f \) is the degrees of freedom (for a monoatomic gas, \( f = 3 \)),
- \( N \) is the number of moles,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Calculate the initial internal energy
For one mole of a monoatomic gas at \( T = 300 \, K \):
\[ U_i = \frac{3}{2} \times 1 \times R \times 300 \]
\[ U_i = \frac{3}{2} \times 300 R = 450 R \]

### Step 3: Determine the final temperature after doubling the volume
Since the pressure is constant and the volume is doubled, we can use the ideal gas law:
\[ \frac{V_f}{V_i} = \frac{T_f}{T_i} \]
Given that \( V_f = 2 V_i \):
\[ \frac{2 V_i}{V_i} = \frac{T_f}{300} \]
This simplifies to:
\[ 2 = \frac{T_f}{300} \]
Thus, the final temperature \( T_f \) is:
\[ T_f = 2 \times 300 = 600 \, K \]

### Step 4: Calculate the final internal energy
Now, we can calculate the final internal energy using the final temperature:
\[ U_f = \frac{3}{2} \times 1 \times R \times 600 \]
\[ U_f = \frac{3}{2} \times 600 R = 900 R \]

### Step 5: Calculate the change in internal energy
The change in internal energy (\( \Delta U \)) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the values we calculated:
\[ \Delta U = 900 R - 450 R = 450 R \]

### Final Answer
The change in internal energy is:
\[ \Delta U = 450 R \]

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