A fixed mass of an ideal gas of volume 50 litre measured at 2 atm and `0^(@)C`. At the same temperature but at 5 atm its volume will be

Options

A. 20 litres
B. 40 litres
C. 60 litres
D. 80 litres

Answer

To solve the problem, we will use Boyle's Law, which states that for a given mass of an ideal gas at constant temperature, the product of pressure and volume is constant. This can be expressed mathematically as:

\[ P_1 V_1 = P_2 V_2 \]

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

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Initial volume, \( V_1 = 50 \, \text{L} \)
- Initial pressure, \( P_1 = 2 \, \text{atm} \)
- Final pressure, \( P_2 = 5 \, \text{atm} \)

**Step 2: Write down Boyle's Law.**
Using Boyle's Law, we have:
\[ P_1 V_1 = P_2 V_2 \]

**Step 3: Substitute the known values into the equation.**
Substituting the values we have:
\[ 2 \, \text{atm} \times 50 \, \text{L} = 5 \, \text{atm} \times V_2 \]

**Step 4: Calculate the left side of the equation.**
Calculating the left side:
\[ 2 \times 50 = 100 \]
So we have:
\[ 100 = 5 \times V_2 \]

**Step 5: Solve for \( V_2 \).**
To find \( V_2 \), we rearrange the equation:
\[ V_2 = \frac{100}{5} \]
Calculating this gives:
\[ V_2 = 20 \, \text{L} \]

### Final Answer:
The volume of the gas at 5 atm and the same temperature will be **20 liters**.

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