A perfect gas goes from a state A to another state B by absorbing 8 × 105 J of heat and doing 6.5 × 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In the second process work done by the gas is ?

Answer

To solve the problem, we will use the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

Where:
- \(\Delta U\) = change in internal energy
- \(Q\) = heat absorbed by the system
- \(W\) = work done by the system

### Step 1: Analyze the first process (from state A to state B)

In the first process:
- Heat absorbed, \(Q_1 = 8 \times 10^5 \, \text{J}\)
- Work done by the gas, \(W_1 = 6.5 \times 10^5 \, \text{J}\)

Using the first law of thermodynamics, we can calculate the change in internal energy (\(\Delta U_1\)) for the first process:

\[ \Delta U_1 = Q_1 - W_1 \]

Substituting the values:

\[ \Delta U_1 = (8 \times 10^5) - (6.5 \times 10^5) \]

Calculating:

\[ \Delta U_1 = 1.5 \times 10^5 \, \text{J} \]

### Step 2: Analyze the second process (from state A to state B)

In the second process:
- Heat absorbed, \(Q_2 = 10^5 \, \text{J}\)
- Work done by the gas, \(W_2 = ?\) (we need to find this)

Since the change in internal energy for the same transition (from state A to state B) must be the same, we have:

\[ \Delta U_2 = \Delta U_1 = 1.5 \times 10^5 \, \text{J} \]

Using the first law of thermodynamics for the second process:

\[ \Delta U_2 = Q_2 - W_2 \]

Substituting the known values:

\[ 1.5 \times 10^5 = 10^5 - W_2 \]

### Step 3: Solve for \(W_2\)

Rearranging the equation to find \(W_2\):

\[ W_2 = 10^5 - 1.5 \times 10^5 \]

Calculating:

\[ W_2 = 10^5 - 1.5 \times 10^5 = -0.5 \times 10^5 \, \text{J} \]

### Final Answer:

The work done by the gas in the second process is:

\[ W_2 = -0.5 \times 10^5 \, \text{J} \]

This negative sign indicates that work is done on the gas.

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