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. a.179 K
B. b.189 K
C. c.199 K
D. d.219 K

Explanation

To find the final temperature of the gas after it expands adiabatically, we can follow these steps:

### Step 1: Identify the given values
- Number of moles (n) = 2 moles
- Initial volume (V1) = 2V
- Final volume (V2) = 4V
- Initial temperature (T1) = 300 K
- For a monoatomic ideal gas, the specific heat ratio (γ) = Cp/Cv = 5/3

### Step 2: Use the adiabatic process equation
For an adiabatic process, we know that:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

### Step 3: Substitute the known values into the equation
We can rearrange the equation to solve for the final temperature (T2):
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

### Step 4: Substitute the values of T1, V1, V2, and γ
Substituting the known values:
- T1 = 300 K
- V1 = 2V
- V2 = 4V
- γ = 5/3

We get:
\[ T_2 = 300 \left( \frac{2V}{4V} \right)^{\frac{5}{3} - 1} \]

### Step 5: Simplify the expression
\[ T_2 = 300 \left( \frac{1}{2} \right)^{\frac{2}{3}} \]

### Step 6: Calculate the final temperature
Now we need to calculate \( \left( \frac{1}{2} \right)^{\frac{2}{3}} \):
1. Calculate \( \frac{1}{2} \) raised to \( \frac{2}{3} \):
\[ \left( \frac{1}{2} \right)^{\frac{2}{3}} = \frac{1}{\sqrt[3]{4}} \]
This is approximately \( 0.7937 \).

2. Now multiply by 300 K:
\[ T_2 = 300 \times 0.7937 \approx 238.11 K \]

### Final Answer
Thus, the final temperature \( T_2 \) is approximately **238 K**.