A perfect gas at `27^(@)C` is heated at constant pressure so as to duble its volume. The tmemperature of th gas will be

Options

A. `300^(@)C`
B. `200^(@)C`
C. `327^(@)C`
D. `600^(@)C`

Answer

To solve the problem, we will use Charles's Law, which states that for a given mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature (in Kelvin).

### Step-by-Step Solution:

1. **Convert the Initial Temperature to Kelvin:**
- The initial temperature \( T_1 \) is given as \( 27^\circ C \).
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore,
\[ T_1 = 27 + 273 = 300 \, K \]

**Hint:** Remember to always convert Celsius to Kelvin when using gas laws.

2. **Identify the Initial and Final Volumes:**
- Let the initial volume \( V_1 = V \).
- Since the volume is doubled, the final volume \( V_2 = 2V \).

**Hint:** Keep track of how the volume changes; in this case, it doubles.

3. **Apply Charles's Law:**
- According to Charles's Law:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Substituting the known values:
\[ \frac{V}{300} = \frac{2V}{T_2} \]

**Hint:** Charles's Law relates volume and temperature; ensure you set up the equation correctly.

4. **Solve for the Final Temperature \( T_2 \):**
- Cross-multiplying gives:
\[ V \cdot T_2 = 2V \cdot 300 \]
- Simplifying (dividing both sides by \( V \)):
\[ T_2 = 2 \cdot 300 = 600 \, K \]

**Hint:** When solving for \( T_2 \), remember to isolate it on one side of the equation.

5. **Convert the Final Temperature Back to Celsius:**
- To convert Kelvin back to Celsius:
\[ T(°C) = T(K) - 273 \]
- Therefore,
\[ T_2(°C) = 600 - 273 = 327^\circ C \]

**Hint:** Always convert back to Celsius if the question asks for the temperature in that unit.

### Final Answer:
The final temperature of the gas will be \( 327^\circ C \).