Problem Statement
Solve the thermodynamics problem: Find the density of nitrogen at $p = 1.0\ \text{atm}$, $T = 300\ \text{K}$. From $pV = \nu RT$ and $\nu = m/M$: $p = \rho RT/M$, so: $$\rho = \frac{pM}{RT} = \frac{1.013\times10^5 \times 0.028}{8.314\times300} = \frac{2836}{2494} \approx 1.14\ \text{kg/m}^3$$ Result: $\rho \approx 1.14\ \text{kg/m}^
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
$$\rho = \frac{pM}{RT} = \frac{1.013\times10^5 \times 0.028}{8.314\times300} = \frac{2836}{2494} \approx 1.14\ \text{kg/m}^3$$
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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