Two moles of an ideal monoatomic gas occupy a volume 2V at temperature 300K, it expands to a volume 4V adiabatically, then the final temperature of gas is

Options

A. 179 K
B. 189 K
C. 199 K
D. 219 K

Explanation

To find the final temperature \( T_2 \) of the gas after it expands adiabatically from volume \( V_1 = 2V \) to volume \( V_2 = 4V \), we can use the adiabatic relation for an ideal gas.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume, \( V_1 = 2V \)
- Final volume, \( V_2 = 4V \)
- Initial temperature, \( T_1 = 300 \, K \)
- Number of moles, \( n = 2 \)
- For a monoatomic ideal gas, the heat capacity ratio \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \).

2. **Use the Adiabatic Relation:**
The relation for adiabatic processes is given by:
\[ \frac{T_1 V_1^{\gamma - 1}}{T_2 V_2^{\gamma - 1}} = 1 \]
Rearranging this gives:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

3. **Substitute the Known Values:**
Substitute \( T_1 \), \( V_1 \), \( V_2 \), and \( \gamma \) into the equation:
\[ T_2 = 300 \left( \frac{2V}{4V} \right)^{\frac{5}{3} - 1} \]

4. **Simplify the Volume Ratio:**
The volume ratio simplifies to:
\[ \frac{2V}{4V} = \frac{1}{2} \]
Therefore:
\[ T_2 = 300 \left( \frac{1}{2} \right)^{\frac{2}{3}} \]

5. **Calculate \( \left( \frac{1}{2} \right)^{\frac{2}{3}} \):**
The value of \( \left( \frac{1}{2} \right)^{\frac{2}{3}} \) can be calculated as:
\[ \left( \frac{1}{2} \right)^{\frac{2}{3}} = \frac{1}{\sqrt[3]{4}} \approx 0.7937 \]

6. **Final Calculation of \( T_2 \):**
Now substitute this value back into the equation for \( T_2 \):
\[ T_2 \approx 300 \times 0.7937 \approx 238.11 \, K \]

7. **Round to Appropriate Significant Figures:**
Rounding this value gives:
\[ T_2 \approx 238 \, K \]

### Final Answer:
The final temperature \( T_2 \) of the gas after adiabatic expansion is approximately **238 K**.

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