A gas undergoes a change of state during which 100J of heat is supplied to it and it does 20J of work. The system is brough back to its original state through a process during which 20 J of heat is released by the gas. What is the work done by the gas in the second process?

Options

A. 60 J
B. 40 J
C. 80 J
D. 20 J

Answer

To solve the problem, we will apply the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). This can be expressed as:

\[ \Delta Q = \Delta U + \Delta W \]

### Step-by-Step Solution:

1. **Identify the Given Values for the First Process:**
- Heat supplied to the gas (Q1) = +100 J (since heat is supplied, it is positive)
- Work done by the gas (W1) = +20 J (work done by the system is positive)

2. **Calculate the Change in Internal Energy (ΔU1) for the First Process:**
Using the first law of thermodynamics:
\[ Q1 = \Delta U1 + W1 \]
Substituting the known values:
\[ 100 = \Delta U1 + 20 \]
Rearranging gives:
\[ \Delta U1 = 100 - 20 = 80 \text{ J} \]

3. **Identify the Given Values for the Second Process:**
- Heat released by the gas (Q2) = -20 J (since heat is released, it is negative)
- Change in internal energy (ΔU2) will be the same in magnitude but opposite in sign to ΔU1, so:
\[ \Delta U2 = -80 \text{ J} \]

4. **Apply the First Law of Thermodynamics for the Second Process:**
Using the first law again:
\[ Q2 = \Delta U2 + W2 \]
Substituting the known values:
\[ -20 = -80 + W2 \]
Rearranging gives:
\[ W2 = -20 + 80 = 60 \text{ J} \]

5. **Conclusion:**
The work done by the gas in the second process is **60 J**.