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 c.c.
B. 510 c.c.
C. 310 c.c.
D. 350 c.c.

Answer

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

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

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

### Step 1: Convert Temperatures to Kelvin
We need to convert the given temperatures from Celsius to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]

- For \( T_1 = 47°C \):
\[ T_1 = 47 + 273.15 = 320.15 \, K \]

- For \( T_2 = 27°C \):
\[ T_2 = 27 + 273.15 = 300.15 \, K \]

### Step 2: Write Down Known Values
Now we can write down the known values:
- \( P_1 = 640 \, mmHg \)
- \( V_1 = ? \) (this is what we need to find)
- \( T_1 = 320.15 \, K \)
- \( P_2 = 620 \, mmHg \)
- \( V_2 = 300 \, cm^3 \)
- \( T_2 = 300.15 \, K \)

### Step 3: Substitute Values into the Combined Gas Law
We will substitute the known values into the combined gas law equation:
\[ \frac{640 \, V_1}{320.15} = \frac{620 \times 300}{300.15} \]

### Step 4: Solve for \( V_1 \)
First, calculate the right side of the equation:
\[ \frac{620 \times 300}{300.15} \approx 619.4 \]

Now, we have:
\[ \frac{640 \, V_1}{320.15} = 619.4 \]

Now, cross-multiply to solve for \( V_1 \):
\[ 640 \, V_1 = 619.4 \times 320.15 \]
\[ V_1 = \frac{619.4 \times 320.15}{640} \]
Calculating the right side:
\[ V_1 \approx \frac{198,750.81}{640} \approx 310.11 \, cm^3 \]

### Final Answer
Thus, the volume of the gas at \( 47°C \) and \( 640 \, mmHg \) is approximately:
\[ V_1 \approx 310 \, cm^3 \]