Two litres of an ideal gas at a pressure of 10 atm expands isothermally at `25^(@)C` into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?

Options

A. 10 J, 10 J
B. 8 J, 10 J
C. 18 J, 0 J
D. 0 J, 0 J

Explanation

To solve the problem, we need to determine the heat absorbed (Q) and the work done (W) during the isothermal expansion of an ideal gas. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify Given Values
- Initial volume (V1) = 2 liters
- Final volume (V2) = 10 liters
- Initial pressure (P1) = 10 atm
- External pressure (P_external) = 0 atm (since it expands into a vacuum)
- Temperature (T) = 25°C = 298 K

### Step 2: Understand the Process
The gas expands isothermally into a vacuum. In an isothermal process, the temperature remains constant, which implies that the change in internal energy (ΔU) is zero for an ideal gas.

### Step 3: Calculate Change in Internal Energy
For an ideal gas, the change in internal energy (ΔU) is given by:
\[ \Delta U = n C_v \Delta T \]
Since the temperature is constant (ΔT = 0), we have:
\[ \Delta U = 0 \]

### Step 4: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = Q + W \]
Substituting ΔU = 0 into the equation gives:
\[ 0 = Q + W \]
This implies:
\[ Q = -W \]

### Step 5: Calculate Work Done (W)
The work done during expansion is given by:
\[ W = -P_{external} \Delta V \]
Where:
\[ \Delta V = V2 - V1 = 10 \, \text{liters} - 2 \, \text{liters} = 8 \, \text{liters} \]
Since the external pressure (P_external) is 0 atm (due to expansion into a vacuum):
\[ W = -0 \times 8 \, \text{liters} = 0 \]

### Step 6: Calculate Heat Absorbed (Q)
Since we found that:
\[ Q = -W \]
And since W = 0:
\[ Q = -0 = 0 \]

### Final Results
- Heat absorbed (Q) = 0 Joules
- Work done (W) = 0 Joules

### Conclusion
In this isothermal expansion into a vacuum, both the heat absorbed and the work done are zero.

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