A gas at temperature `270^(0)C` and pressure `30` atmosphere is allowed to expand to one atmospheric pressure. If the volume, the final temerpature becomes

Options

A. `100^(0)C`
B. `373^(0)K`
C. `373^(0)C`
D. `-173^(0)C`

Explanation

To solve the problem, we will use the ideal gas law and the relationship between pressure, volume, and temperature for a gas undergoing expansion.

### Step-by-Step Solution:

1. **Convert Initial Temperature to Kelvin**:
The initial temperature \( T_1 \) is given as \( 27^{\circ}C \). To convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

2. **Identify Initial Conditions**:
The initial pressure \( P_1 \) is \( 30 \, atm \) and the initial volume \( V_1 \) is \( V \).

3. **Identify Final Conditions**:
The final pressure \( P_2 \) is \( 1 \, atm \) and the final volume \( V_2 \) is \( 10V \) (the volume expands to 10 times the initial volume).

4. **Use the Ideal Gas Law Relationship**:
According to the ideal gas law, we can use the relationship:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Plugging in the known values:
\[ \frac{30 \, atm \cdot V}{300 \, K} = \frac{1 \, atm \cdot 10V}{T_2} \]

5. **Simplify the Equation**:
The volume \( V \) cancels out from both sides:
\[ \frac{30}{300} = \frac{10}{T_2} \]
This simplifies to:
\[ \frac{1}{10} = \frac{10}{T_2} \]

6. **Cross-Multiply to Solve for \( T_2 \)**:
Cross-multiplying gives:
\[ T_2 = 10 \times 10 = 100 \, K \]

7. **Convert Final Temperature to Celsius**:
To convert \( T_2 \) back to Celsius:
\[ T_2 = 100 \, K - 273 = -173^{\circ}C \]

### Final Answer:
The final temperature \( T_2 \) is \( -173^{\circ}C \).

---