Question

A gas is expanded from an initial state to a final state along a path that consists of (a) an isothermal expansion doing 40 J work, (b) an adiabatic expansion doing W work, (c) an isothermal expansion doing 30 J work. If the total change in the internal energy of the gas is -20 J, the work done by the gas during the adiabatic expansion W =

• A. $50 \text{ J}$
• B. $90 \text{ J}$
• C. $70 \text{ J}$
• D. $20 \text{ J}$

Hint:

Work \& Energy: For isothermal: $\Delta U = 0$
For adiabatic: $\Delta Q = 0 \Rightarrow \Delta U = -W$
Total: $\Delta U_\texttotal = \sum \Delta U_i$

Answer 1

The Correct Option is D

In the first and third steps (isothermal), the internal energy does not change: \[ \Delta U_{1} = \Delta U_{3} = 0 \] Let the work done during the adiabatic process be $W$. For adiabatic: \[ \Delta U_{2} = -W \] The total internal energy change: \[ \Delta U = \Delta U_1 + \Delta U_2 + \Delta U_3 = -W \] Given $\Delta U = -20$ J, hence: \[ -W = -20 \Rightarrow W = 20~\text{J} \] So, the adiabatic work done is $20~\text{J}$, matching option (4).

Answer 2

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<b>Step 1: Understand the problem and given data</b><br>  
- The gas undergoes three expansions:<br>  
 (a) Isothermal expansion doing 40 J work<br>  
 (b) Adiabatic expansion doing W work (unknown)<br>  
 (c) Isothermal expansion doing 30 J work<br>  
- Total change in internal energy, ΔU = -20 J<br><br>

<b>Step 2: Recall key concepts about work and internal energy</b><br>  
- For an isothermal process in an ideal gas, internal energy change ΔU = 0 (since temperature is constant).<br>  
- For an adiabatic process, the change in internal energy equals the work done by the gas:<br>  
 ΔU_adiabatic = W_adiabatic (work done by the gas).<br><br>

<b>Step 3: Calculate total work done and relate to internal energy</b><br>  
Total work done by the gas is:<br>  
W_total = 40 J + W + 30 J = (70 + W) J<br><br>

The total change in internal energy is the sum of changes in all steps:<br>  
ΔU_total = ΔU_isothermal1 + ΔU_adiabatic + ΔU_isothermal2<br>  
Since isothermal ΔU = 0, we get:<br>  
ΔU_total = ΔU_adiabatic = W<br><br>

<b>Step 4: Use the given ΔU to find W</b><br>  
Given ΔU_total = -20 J,<br>  
So, W = ΔU_total = -20 J<br>  

However, this contradicts the idea that W is the work done by the gas during adiabatic expansion (work done by gas should be positive for expansion). So we need to carefully analyze.<br><br>

<b>Step 5: Use the first law of thermodynamics</b><br>  
First law: ΔU = Q - W_total<br>  
Rearranged: W_total = Q - ΔU<br>  
Since the total work is W_total = 70 + W,<br>  
We need to find W.<br><br>

<b>Step 6: Consider heat exchange</b><br>  
- For isothermal expansion, heat absorbed Q equals work done:<br>  
Q_isothermal1 = 40 J<br>  
Q_isothermal2 = 30 J<br>  
- For adiabatic expansion, Q_adiabatic = 0 (no heat exchange).<br>  
So total heat absorbed:<br>  
Q_total = 40 + 0 + 30 = 70 J<br><br>

<b>Step 7: Calculate W using first law</b><br>  
ΔU = Q_total - W_total<br>  
-20 = 70 - (70 + W)<br>  
-20 = 70 - 70 - W<br>  
-20 = -W<br>  
W = 20 J<br><br>

<b>Step 8: Conclusion</b><br>  
The work done by the gas during the adiabatic expansion is 20 J.<br>
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