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

Options

A. `1^(@)C`
B. `900^(@)C`
C. `627^(@)C`
D. `450^(@)C`

Answer

To solve the problem, we need to determine the final temperature of a perfect gas that is heated at constant pressure, resulting in a tripling of its volume. Here’s a step-by-step breakdown of the solution:

### Step 1: Convert the initial temperature from Celsius to Kelvin
The initial temperature \( T_1 \) is given as \( 27^\circ C \). To convert this to Kelvin, we use the formula:
\[ T_1 = 27 + 273 = 300 \, K \]

### Step 2: Define the initial and final volumes
Let the initial volume \( V_1 \) be \( V \). Since the volume triples, the final volume \( V_2 \) will be:
\[ V_2 = 3V \]

### Step 3: Use the ideal gas law relationship
Since the pressure is constant, we can use the relationship from the ideal gas law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Given that \( P_1 = P_2 = P \) (constant pressure), we can simplify this to:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

### Step 4: Substitute the known values
Substituting \( V_1 = V \), \( V_2 = 3V \), and \( T_1 = 300 \, K \) into the equation gives:
\[ \frac{V}{300} = \frac{3V}{T_2} \]

### Step 5: Cancel out the common terms
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{1}{300} = \frac{3}{T_2} \]

### Step 6: Solve for \( T_2 \)
Cross-multiplying gives:
\[ T_2 = 3 \times 300 = 900 \, K \]

### Step 7: Convert the final temperature from Kelvin to Celsius
To convert \( T_2 \) back to Celsius, we use:
\[ T_2 (^\circ C) = T_2 (K) - 273 = 900 - 273 = 627 \, ^\circ C \]

### Final Answer
The final temperature of the gas will be \( 627 \, ^\circ C \).

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