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. `20 xx (1)/(50) xx 2`
D. 50

Answer

To solve the problem of finding the final pressure of a gas when it expands from 20 cm³ to 50 cm³ at constant temperature, we can use Boyle's Law. Boyle's Law states that the pressure of a gas is inversely proportional to its volume when the temperature is held constant. Mathematically, this is expressed 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:

1. **Identify the Given Values:**
- Initial pressure, \( P_1 = 1 \, \text{atm} \)
- Initial volume, \( V_1 = 20 \, \text{cm}^3 \)
- Final volume, \( V_2 = 50 \, \text{cm}^3 \)
- Final pressure, \( P_2 = ? \)

2. **Apply Boyle's Law:**
Using the formula \( P_1 V_1 = P_2 V_2 \), we can rearrange it to solve for \( P_2 \):

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

3. **Substitute the Known Values:**
Plugging in the values we have:

\[ P_2 = \frac{(1 \, \text{atm}) \times (20 \, \text{cm}^3)}{50 \, \text{cm}^3} \]

4. **Calculate \( P_2 \):**
Now, perform the calculation:

\[ P_2 = \frac{20 \, \text{atm} \cdot \text{cm}^3}{50 \, \text{cm}^3} = \frac{20}{50} \, \text{atm} = 0.4 \, \text{atm} \]

5. **Final Result:**
The final pressure \( P_2 \) after the gas expands to 50 cm³ at constant temperature is:

\[ P_2 = 0.4 \, \text{atm} \]