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 during an isothermal compression from 30 liters to 6 liters at STP, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial volume (V1) = 30 liters
- Final volume (V2) = 6 liters
- Standard Temperature (T) = 0°C = 273 K
- Standard Pressure (P) = 1 atm (not directly needed for calculations)
- Gas constant (R) = 8.314 J/(mol·K)

2. **Calculate the Number of Moles (n):**
The number of moles of the gas at STP can be calculated using the formula:
\[ n = \frac{V}{22.4} \]
For V1 = 30 liters:
\[ n = \frac{30}{22.4} \approx 1.339 \text{ moles} \]

3. **Calculate the Work Done (W) in the Isothermal Process:**
The work done during an isothermal process is given by the formula:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]
Substituting the values:
\[ W = 1.339 \times 8.314 \times 273 \times \ln\left(\frac{6}{30}\right) \]
First, calculate \(\ln\left(\frac{6}{30}\right) = \ln(0.2) \approx -1.609\).

Now substituting this back into the work equation:
\[ W = 1.339 \times 8.314 \times 273 \times (-1.609) \]

4. **Calculate the Result:**
Performing the calculations step-by-step:
- Calculate \(1.339 \times 8.314 \approx 11.126\)
- Then, \(11.126 \times 273 \approx 3036.378\)
- Finally, \(3036.378 \times (-1.609) \approx -4882.59 \text{ J}\)

5. **Determine the Heat Extracted (Q):**
According to the first law of thermodynamics for an isothermal process, the heat extracted (Q) is equal to the work done (W):
\[ Q = W \]
Thus, the heat taken out of the gas is:
\[ Q \approx -4882.59 \text{ J} \approx -4.88 \text{ kJ} \]

### Final Answer:
The amount of heat taken out of the gas is approximately \(-4.88 \text{ kJ}\).

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