30 litre of a gas at STP is isothermally compressed to 6 litre. Amount of heat taken out of the gas is:

Options

A. `8.05 j`
B. `4.8 kJ`
C. `100 J`
D. `150 J`

Explanation

To solve the problem of finding the amount of heat taken out of a gas that is isothermally compressed from 30 liters to 6 liters at STP, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**:
- The gas is compressed isothermally, which means the temperature remains constant throughout the process.
- At STP (Standard Temperature and Pressure), the temperature is 273 K.

2. **Identify Key Variables**:
- Initial volume, \( V_1 = 30 \, \text{liters} \)
- Final volume, \( V_2 = 6 \, \text{liters} \)
- The gas constant, \( R = 8.314 \, \text{J/(mol·K)} \)

3. **Calculate the Number of Moles**:
- At STP, 1 mole of an ideal gas occupies 22.4 liters.
- Therefore, the number of moles \( n \) can be calculated as:
\[ n = \frac{30 \, \text{liters}}{22.4 \, \text{liters/mole}} \approx 1.34 \, \text{moles} \]

4. **Apply the First Law of Thermodynamics**:
- The first law states:
\[ q = \Delta U + W \]
- For an isothermal process involving an ideal gas, the change in internal energy \( \Delta U = 0 \). Thus:
\[ q = W \]

5. **Calculate the Work Done**:
- The work done on the gas during isothermal compression is given by the formula:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]
- Substituting the values:
\[ W = 1.34 \, \text{moles} \times 8.314 \, \text{J/(mol·K)} \times 273 \, \text{K} \times \ln\left(\frac{6}{30}\right) \]

6. **Calculate the Natural Logarithm**:
- Calculate \( \ln\left(\frac{6}{30}\right) = \ln(0.2) \).
- This value is approximately -1.609.

7. **Substitute and Calculate Work**:
- Now substituting all the values:
\[ W = 1.34 \times 8.314 \times 273 \times (-1.609) \]
- Performing the calculations:
\[ W \approx 1.34 \times 8.314 \times 273 \times -1.609 \approx -4.8 \, \text{kJ} \]

8. **Determine Heat Taken Out**:
- Since \( q = W \), the amount of heat taken out of the gas is:
\[ q \approx -4.8 \, \text{kJ} \]
- The negative sign indicates that heat is being removed from the gas.

### Final Answer:
The amount of heat taken out of the gas is approximately **-4.8 kJ**.