1 mole of gas expands isothermally at `37^(@)C`. The amount of heat is absorbed by it until its volume doubled is `(R= 8.31 J mol^(-1) K^(-1))`

Options

A. a. 411.25 cal
B. b. 418.50 cal
C. c. 420.25 cal
D. d. 425.40 cal

Explanation

To solve the problem of calculating the amount of heat absorbed by 1 mole of gas during an isothermal expansion at a temperature of 37°C until its volume doubles, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin:**
The temperature given is in Celsius. To convert it to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
For \(37°C\):
\[ T = 37 + 273.15 = 310.15 \, K \approx 310 \, K \]

2. **Identify Given Values:**
- Number of moles, \(n = 1 \, \text{mol}\)
- Universal gas constant, \(R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}\)
- Initial volume, \(V_1\)
- Final volume, \(V_2 = 2V_1\)

3. **Use the Formula for Heat Change in Isothermal Process:**
The heat absorbed (\(\Delta Q\)) during an isothermal expansion is given by:
\[ \Delta Q = nRT \ln\left(\frac{V_2}{V_1}\right) \]
Substituting the known values:
\[ \Delta Q = 1 \times 8.31 \times 310 \times \ln\left(\frac{2V_1}{V_1}\right) \]
Simplifying the logarithm:
\[ \Delta Q = 8.31 \times 310 \times \ln(2) \]

4. **Calculate \(\ln(2)\):**
The natural logarithm of 2 is approximately:
\[ \ln(2) \approx 0.693 \]

5. **Calculate \(\Delta Q\):**
Now substituting \(\ln(2)\) into the equation:
\[ \Delta Q = 8.31 \times 310 \times 0.693 \]
Performing the multiplication:
\[ \Delta Q \approx 8.31 \times 310 \times 0.693 \approx 785.61 \, \text{J} \]

6. **Convert Joules to Calories:**
To convert the heat from joules to calories, we use the conversion factor \(1 \, \text{cal} = 4.2 \, \text{J}\):
\[ Q(\text{cal}) = \frac{785.61 \, \text{J}}{4.2 \, \text{J/cal}} \approx 186.76 \, \text{cal} \]

### Final Answer:
The amount of heat absorbed by the gas is approximately \(186.76 \, \text{cal}\).