Problem Statement
Solve the thermodynamics problem: One mole of monatomic ideal gas is heated at constant pressure from $T_1=300\ \text{K}$ to $T_2=600\ \text{K}$. Find $W$, $\Delta U$, and $Q$. Work at constant pressure: $W = \nu R\Delta T = 1.0\times8.314\times300 = 2494\ \text{J} \approx 2.5\ \text{kJ}$. Internal energy change (monatomic: $C_v = \
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Thermodynamics governs energy transformations involving heat and work. The First Law $\Delta U = Q – W$ expresses energy conservation. For an ideal gas, internal energy depends only on temperature ($U = nC_VT$), and the equation of state $PV = nRT$ links pressure, volume, and temperature.
- $\Delta U = Q – W$ — First Law of Thermodynamics
- $PV = nRT$ — ideal gas equation
- $C_P – C_V = R$, $\gamma = C_P/C_V$
- $W = \int P\,dV$ — work done by gas
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
Full substitution shown in the steps above.
Answer
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
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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