A gas occupies a volume of 300 `cm^(3)` at `27 .^(@)C` and 620 mm pressure . The volume of gas at `47 .^(@)C` and 640 mm pressure is

Options

A. 400 cc
B. 510 cc
C. 310 cc
D. 350 cc

Answer

To solve the problem, we will use the combined gas law, which relates pressure, volume, and temperature of a gas. The formula is:

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial Pressure, \( P_1 = 620 \, \text{mm Hg} \)
- Initial Volume, \( V_1 = 300 \, \text{cm}^3 \)
- Initial Temperature, \( T_1 = 27 \, \text{°C} \)
- Final Pressure, \( P_2 = 640 \, \text{mm Hg} \)
- Final Temperature, \( T_2 = 47 \, \text{°C} \)

2. **Convert Temperatures to Kelvin:**
- For \( T_1 \):
\[ T_1 = 27 + 273 = 300 \, \text{K} \]
- For \( T_2 \):
\[ T_2 = 47 + 273 = 320 \, \text{K} \]

3. **Set Up the Equation Using the Combined Gas Law:**
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

4. **Substitute the Known Values into the Equation:**
\[ \frac{620 \times 300}{300} = \frac{640 \times V_2}{320} \]

5. **Simplify the Equation:**
- The \( 300 \) in the numerator and denominator cancels out:
\[ 620 = \frac{640 \times V_2}{320} \]

6. **Cross-Multiply to Solve for \( V_2 \):**
\[ 620 \times 320 = 640 \times V_2 \]

7. **Calculate \( 620 \times 320 \):**
\[ 620 \times 320 = 198400 \]

8. **Now, Isolate \( V_2 \):**
\[ V_2 = \frac{198400}{640} \]

9. **Calculate \( V_2 \):**
\[ V_2 = 310 \, \text{cm}^3 \]

### Final Answer:
The volume of the gas at \( 47 \, \text{°C} \) and \( 640 \, \text{mm} \) pressure is \( \boxed{310 \, \text{cm}^3} \).