Work done by 0.1 mole of a gas at `27^@ C` to double its volume at constant pressure is (use R=2 cal/mol`""^(@)`C)

Options

A. 54 cal
B. 600 cal
C. 60 cal
D. 546 cal

Answer

To solve the problem of calculating the work done by 0.1 mole of a gas at 27°C to double its volume at constant pressure, we will follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Celsius is given as 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
So,
\[ T_1 = 27 + 273 = 300 \, K \]

### Step 2: Determine the initial and final volumes
Let the initial volume be \( V \). Since the volume is doubled, the final volume \( V_2 \) will be:
\[ V_2 = 2V \]

### Step 3: Use the relationship between volume and temperature at constant pressure
According to Charles's law, at constant pressure, the ratio of volume to temperature is constant:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Substituting the known values:
\[ \frac{V}{300} = \frac{2V}{T_2} \]
Cancelling \( V \) from both sides:
\[ \frac{1}{300} = \frac{2}{T_2} \]
Cross-multiplying gives:
\[ T_2 = 600 \, K \]

### Step 4: Calculate the change in temperature
The change in temperature \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 = 600 - 300 = 300 \, K \]

### Step 5: Calculate the work done using the formula
The work done \( W \) at constant pressure can be calculated using the formula:
\[ W = nR\Delta T \]
Where:
- \( n = 0.1 \, \text{moles} \)
- \( R = 2 \, \text{cal/mol} \cdot °C \) (Note: 1 cal = 4.184 J, so we will convert calories to joules)
- \( \Delta T = 300 \, K \)

Substituting the values:
\[ W = 0.1 \times 2 \times 300 \]
\[ W = 60 \, \text{cal} \]

### Step 6: Convert the work done from calories to joules
To convert calories to joules:
\[ W = 60 \, \text{cal} \times 4.184 \, \text{J/cal} = 251.04 \, \text{J} \]

### Final Answer
The work done by the gas is approximately:
\[ W \approx 251.04 \, \text{J} \]