The raise in the temperature of a given mass of an ideal gas at constant pressure and at temperature `27^@` to double its volume is

Options

A. `327^@ C`
B. `54^@ C`
C. `300^@ C`
D. `600^@ C`

Answer

To solve the problem of finding the rise in temperature of a given mass of an ideal gas at constant pressure when its volume is doubled, we can follow these steps:

### Step 1: Understand the relationship between volume and temperature
For an ideal gas at constant pressure, the volume (V) is directly proportional to the absolute temperature (T) according to Charles's Law. This can be expressed as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
where \( V_1 \) and \( T_1 \) are the initial volume and temperature, and \( V_2 \) and \( T_2 \) are the final volume and temperature.

### Step 2: Convert the initial temperature to Kelvin
The initial temperature given is \( 27^\circ C \). To convert this to Kelvin, we add 273:
\[ T_1 = 27 + 273 = 300 \, K \]

### Step 3: Define the initial and final volumes
Let the initial volume be \( V_1 = V \) and the final volume be \( V_2 = 2V \) (since the volume is doubled).

### Step 4: Set up the equation using Charles's Law
Using the relationship from Step 1:
\[ \frac{V}{T_1} = \frac{2V}{T_2} \]

### Step 5: Simplify the equation
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{1}{T_1} = \frac{2}{T_2} \]

### Step 6: Rearrange to find \( T_2 \)
Rearranging gives:
\[ T_2 = 2T_1 \]

### Step 7: Substitute the value of \( T_1 \)
Substituting \( T_1 = 300 \, K \):
\[ T_2 = 2 \times 300 = 600 \, K \]

### Step 8: Convert the final temperature back to Celsius
To convert \( T_2 \) back to Celsius:
\[ T_2 = 600 - 273 = 327^\circ C \]

### Step 9: Calculate the rise in temperature
The rise in temperature \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 \]
Substituting the values:
\[ \Delta T = 327 - 27 = 300^\circ C \]

### Final Answer
The rise in temperature is \( 300^\circ C \).

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