If `20 cm^(3)` gas at `1 atm` is expanded to `50 cm^(3)` at constant `T`, then what is the final pressure

Options

A. `20 xx 1/50 `
B. `50 XX 1/20`
C. `1 XX 1/20 XX 50 `
D. None of these

Answer

To solve the problem of finding the final pressure (P2) after a gas expands from 20 cm³ to 50 cm³ at constant temperature, we will use Boyle's Law. Boyle's Law states that for a given mass of gas at constant temperature, the product of pressure and volume is constant. This can be mathematically expressed as:

\[ P_1 V_1 = P_2 V_2 \]

**Step-by-step solution:**

1. **Identify the given values:**
- Initial volume (V1) = 20 cm³
- Final volume (V2) = 50 cm³
- Initial pressure (P1) = 1 atm
- Final pressure (P2) = ?

2. **Write down Boyle's Law:**
\[ P_1 V_1 = P_2 V_2 \]

3. **Substitute the known values into the equation:**
\[ 1 \, \text{atm} \times 20 \, \text{cm}^3 = P_2 \times 50 \, \text{cm}^3 \]

4. **Rearrange the equation to solve for P2:**
\[ P_2 = \frac{P_1 V_1}{V_2} \]

5. **Substitute the values into the rearranged equation:**
\[ P_2 = \frac{1 \, \text{atm} \times 20 \, \text{cm}^3}{50 \, \text{cm}^3} \]

6. **Calculate P2:**
\[ P_2 = \frac{20}{50} \, \text{atm} \]
\[ P_2 = 0.4 \, \text{atm} \]

Thus, the final pressure (P2) after the gas expands to 50 cm³ at constant temperature is **0.4 atm**.