How much energy is absorbed by 10 moles of an ideal gas if it expands from an initial pressure of 8 atmosphere to 4 atmosphere at a constant temperature of `27^(@)C`?
`(R=8.31J" "mol^(-1)" "K^(-1))`

Explanation

To solve the problem of how much energy is absorbed by 10 moles of an ideal gas when it expands from an initial pressure of 8 atm to 4 atm at a constant temperature of 27°C, we can follow these steps:

### Step 1: Understand the System
We are dealing with an ideal gas that is expanding isothermally (constant temperature). The temperature is given as 27°C, which we need to convert to Kelvin.

**Hint:** Remember to convert Celsius to Kelvin by adding 273.15.

### Step 2: Convert Temperature to Kelvin
\[ T = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 3: Identify the Initial and Final Pressures
The initial pressure \( P_1 \) is 8 atm and the final pressure \( P_2 \) is 4 atm.

### Step 4: Use the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = Q + W \]
For an isothermal process involving an ideal gas, the change in internal energy (\( \Delta U \)) is zero:
\[ \Delta U = 0 \implies Q = -W \]

### Step 5: Calculate Work Done (W)
The work done by the gas during an isothermal expansion can be calculated using the formula:
\[ W = -2.303 \, nRT \log\left(\frac{P_1}{P_2}\right) \]
Where:
- \( n = 10 \) moles
- \( R = 8.314 \, J \, mol^{-1} \, K^{-1} \)
- \( T = 300 \, K \)
- \( P_1 = 8 \, atm \)
- \( P_2 = 4 \, atm \)

### Step 6: Substitute the Values into the Work Formula
\[ W = -2.303 \times 10 \times 8.314 \times 300 \log\left(\frac{8}{4}\right) \]

### Step 7: Calculate the Logarithm
\[ \log\left(\frac{8}{4}\right) = \log(2) \approx 0.301 \]

### Step 8: Substitute the Logarithm Value
\[ W = -2.303 \times 10 \times 8.314 \times 300 \times 0.301 \]

### Step 9: Perform the Calculation
Calculating the values step by step:
1. \( 2.303 \times 10 = 23.03 \)
2. \( 23.03 \times 8.314 \approx 191.1 \)
3. \( 191.1 \times 300 \approx 57330 \)
4. \( 57330 \times 0.301 \approx 17240.93 \)

Thus,
\[ W \approx -17240.93 \, J \]

### Step 10: Calculate Heat Absorbed (Q)
Since \( Q = -W \):
\[ Q = 17240.93 \, J \]

### Final Answer
The energy absorbed by the gas is approximately:
\[ Q \approx 17240.93 \, J \approx 17241 \, J \]

### Summary of Steps
1. Convert temperature to Kelvin.
2. Identify initial and final pressures.
3. Use the first law of thermodynamics.
4. Calculate work done using the isothermal expansion formula.
5. Substitute values and calculate logarithm.
6. Perform the final calculations to find energy absorbed.