How much heat is absorbed if the expansion is carried out from 2.5L to final volume of 12.5L reversibly? (Give your answer in Joule)

Explanation

To solve the problem of how much heat is absorbed during a reversible expansion from a volume of 2.5 L to 12.5 L, we can follow these steps:

### Step 1: Understand the process
We are dealing with a reversible expansion of an ideal gas. The work done during this process can be calculated using the formula for work done in a reversible process.

### Step 2: Write the formula for work done
The work done (W) during a reversible expansion can be expressed as:
\[ W = -nRT \ln \left(\frac{V_2}{V_1}\right) \]
Where:
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = absolute temperature in Kelvin
- \( V_1 \) = initial volume (2.5 L)
- \( V_2 \) = final volume (12.5 L)

### Step 3: Substitute the values
Assuming we have the values for \( n \) (number of moles) and \( T \) (temperature), we can substitute them into the formula. However, since these values are not provided in the question, we will proceed with the calculation assuming \( n \) and \( T \) are known.

### Step 4: Calculate the work done
Using the formula:
\[ W = -nRT \ln \left(\frac{12.5}{2.5}\right) \]
\[ W = -nRT \ln(5) \]

### Step 5: Convert work done to Joules
If we have the pressure of the gas, we can relate the work done to the heat absorbed using the first law of thermodynamics:
\[ Q = W + \Delta U \]
For an ideal gas, if we assume the process is isothermal, then \( \Delta U = 0 \) and thus:
\[ Q = W \]

### Step 6: Substitute pressure and convert units
Assuming the pressure \( P \) is given as 10 atm, we can convert this to Joules:
1 atm = 101.3 J/L
Thus, \( P = 10 \times 101.3 \, \text{J/L} = 1013 \, \text{J/L} \).

### Step 7: Final calculation
Now substituting back into the work done formula:
\[ W = -nRT \ln(5) \]
Assuming \( n = 1 \) mole and \( T = 298 \, K \):
\[ W = -1 \times 8.314 \times 298 \times \ln(5) \]
Calculating \( \ln(5) \approx 1.609 \):
\[ W = -1 \times 8.314 \times 298 \times 1.609 \]
\[ W \approx -3980.9 \, \text{J} \]

### Step 8: Conclusion
Thus, the heat absorbed \( Q \) during the process is approximately:
\[ Q \approx 3980.9 \, \text{J} \]