At what temperature, the sample of neon gas would be heated to double of its pressure, if the initial volume of gas is/are reduced to 15% at 44.4 K

Options

A. `319^(@)C`
B. `592^(@)C`
C. `128^(@)C`
D. `90^(@)C`

Answer

To solve the problem, we will use the ideal gas law, which can be expressed in the form of the equation:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature
- \( P_2 \) = final pressure
- \( V_2 \) = final volume
- \( T_2 \) = final temperature

### Step 1: Identify the known values
- Initial temperature, \( T_1 = 44.4 \, K \)
- Initial pressure, \( P_1 = P \)
- Final pressure, \( P_2 = 2P \) (since the pressure is doubled)
- Initial volume, \( V_1 = V \)
- Final volume, \( V_2 = 15\% \, \text{of} \, V = 0.15V \)

### Step 2: Substitute the known values into the equation
We can rewrite the equation using the known values:

\[ \frac{P \cdot V}{44.4} = \frac{2P \cdot 0.15V}{T_2} \]

### Step 3: Simplify the equation
We can cancel \( P \) and \( V \) from both sides of the equation (assuming they are not zero):

\[ \frac{1}{44.4} = \frac{2 \cdot 0.15}{T_2} \]

### Step 4: Solve for \( T_2 \)
Rearranging the equation to solve for \( T_2 \):

\[ T_2 = 2 \cdot 0.15 \cdot 44.4 \]

Calculating \( T_2 \):

\[ T_2 = 0.3 \cdot 44.4 = 13.32 \, K \]

### Step 5: Convert to Celsius
To convert from Kelvin to Celsius, we use the formula:

\[ T(°C) = T(K) - 273.15 \]

So,

\[ T(°C) = 13.32 - 273.15 = -259.83 °C \]

### Final Answer
The final temperature \( T_2 \) at which the sample of neon gas would be heated to double its pressure is approximately **-259.83 °C**.

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