Problem Statement
An ideal gas undergoes a cycle: isothermal expansion ($1\to2$), isochoric cooling ($2\to3$), adiabatic compression ($3\to1$). Find the net work per cycle in terms of $T_1, T_2, \nu$.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Process 1→2 (isothermal at $T_1$): $W_{12} = \nu RT_1\ln(V_2/V_1)$.
Step 2 — Apply the relevant physical law or equation: Process 2→3 (isochoric): $W_{23} = 0$.
Step 3 — Solve algebraically for the unknown: Process 3→1 (adiabatic): $W_{31} = -\Delta U_{31} = -\nu C_v(T_1-T_2) = \nu C_v(T_2-T_1)$. Wait — gas is compressed from $T_2$ back to $T_1$, so:
Step 4 — Substitute numerical values with units: $$W_{31} = \nu C_v(T_2-T_1)\ \text{(work done by gas, negative if }T_1>T_2)$$
Step 5 — Compute and check the result: Net work:
Step 6: $$W_{net} = \nu RT_1\ln\frac{V_2}{V_1} + \nu C_v(T_2-T_1)$$
Worked Calculation
$$W_{31} = \nu C_v(T_2-T_1)\ \text{(work done by gas, negative if }T_1>T_2)$$
$$W_{net} = \nu RT_1\ln\frac{V_2}{V_1} + \nu C_v(T_2-T_1)$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
Process 1→2 (isothermal at $T_1$): $W_{12} = \nu RT_1\ln(V_2/V_1)$.
Process 2→3 (isochoric): $W_{23} = 0$.
Process 3→1 (adiabatic): $W_{31} = -\Delta U_{31} = -\nu C_v(T_1-T_2) = \nu C_v(T_2-T_1)$. Wait — gas is compressed from $T_2$ back to $T_1$, so:
$$W_{31} = \nu C_v(T_2-T_1)\ \text{(work done by gas, negative if }T_1>T_2)$$
Net work:
$$W_{net} = \nu RT_1\ln\frac{V_2}{V_1} + \nu C_v(T_2-T_1)$$
Since $\Delta U_{cycle}=0$ (state function), $W_{net} = Q_{net} = Q_{12}+Q_{23}$ (only isothermal and isochoric steps exchange heat).
Answer
$$\boxed{W_{net} = \nu RT_1\ln\frac{V_2}{V_1} + \nu C_v(T_2-T_1)}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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