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. ` 260 cm^(3)`
B. `310 cm^(3)`
C. `390 cm^(3)`
D. `450 cm^(3)`

Explanation

To solve the problem of finding the volume of gas at different conditions of temperature and pressure, we can use the combined gas law, which states that:

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

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature (in Kelvin)
- \( P_2 \) = final pressure
- \( V_2 \) = final volume (what we want to find)
- \( T_2 \) = final temperature (in Kelvin)

### Step 1: Convert temperatures from Celsius to Kelvin
- For \( T_1 = 27^\circ C \):
\[ T_1 = 27 + 273 = 300 \, K \]

- For \( T_2 = 47^\circ C \):
\[ T_2 = 47 + 273 = 320 \, K \]

### Step 2: Identify the known values
- \( P_1 = 620 \, mmHg \)
- \( V_1 = 300 \, cm^3 \)
- \( T_1 = 300 \, K \)
- \( P_2 = 640 \, mmHg \)
- \( T_2 = 320 \, K \)

### Step 3: Substitute the known values into the combined gas law
Using the combined gas law:
\[ \frac{620 \, mmHg \times 300 \, cm^3}{300 \, K} = \frac{640 \, mmHg \times V_2}{320 \, K} \]

### Step 4: Simplify the equation
Cross-multiplying gives:
\[ 620 \times 300 \times 320 = 640 \times V_2 \times 300 \]

### Step 5: Solve for \( V_2 \)
Rearranging for \( V_2 \):
\[ V_2 = \frac{620 \times 300 \times 320}{640 \times 300} \]

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

### Step 6: Calculate \( V_2 \)
Calculating the above expression:
\[ V_2 = \frac{620 \times 320}{640} = \frac{198400}{640} = 310 \, cm^3 \]

### Final Answer
The volume of the gas at \( 47^\circ C \) and \( 640 \, mmHg \) is \( 310 \, cm^3 \).

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