Category: Part 2: Thermodynamics
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Problem 2.10 — RMS Speed of Gas Molecules
Problem Statement Find the rms speed of nitrogen molecules at $T = 300\ \text{K}$. 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…
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Problem 2.8 — Loschmidt Number: Molecules per cm³ at STP
Problem Statement Find the number of ideal gas molecules in $1.0\ \text{cm}^3$ at $p = 1.0\ \text{atm}$, $T = 273\ \text{K}$. 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…
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Problem 2.9 — Gas Mixture: Mean Molar Mass
Problem Statement A mixture contains $m_1 = 14\ \text{g}$ of $\text{N}_2$ and $m_2 = 16\ \text{g}$ of $\text{O}_2$. Find the mean molar mass $\bar{M}$. 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…
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Problem 2.6 — Gay-Lussac’s Law: Pressure at Constant Volume
Problem Statement Solve the fluid mechanics problem: A gas is sealed at $p_1 = 1.0\ \text{atm}$, $T_1 = 17°\text{C}$. It is heated to $T_2 = 107°\text{C}$ at constant volume. Find $p_2$. Gay-Lussac’s Law: $p/T = \text{const}$ at constant volume. $$p_2 = p_1\frac{T_2}{T_1} = 1.0\times\frac{380}{290} \approx 1.31\ \text{atm}$$ ($T_1 = 290\ \text{K}$, $T Given Information See…
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Problem 2.7 — Charles’s Law: Volume at Constant Pressure
Problem Statement Solve the fluid mechanics problem: An ideal gas occupies $V_1 = 1.0\ \text{L}$ at $T_1 = 300\ \text{K}$. Find the volume at $T_2 = 450\ \text{K}$ at constant pressure. $$\frac{V_1}{T_1}=\frac{V_2}{T_2} \implies V_2 = V_1\frac{T_2}{T_1} = 1.0\times\frac{450}{300} = 1.5\ \text{L}$$ Result: $V_2 = 1.5\ \text{L}$. Given Information See problem statement for all given quantities.…
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Problem 2.4 — Gas Mixture: Partial Pressures and Total Pressure
Problem Statement Solve the fluid mechanics problem: A vessel of volume $V = 25\ \text{L}$ contains $\nu_1 = 1.0\ \text{mol}$ of $\text{N}_2$ and $\nu_2 = 2.0\ \text{mol}$ of $\text{O}_2$ at $T = 300\ \text{K}$. Find the partial and total pressures. Each gas obeys the ideal gas law independently (Dalton’s law): $$p_i = \frac{\nu_i RT}{V}$$ $$p_{N_2}…
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Problem 2.5 — Ideal Gas Density vs Pressure
Problem Statement Solve the thermodynamics problem: Find the density of nitrogen at $p = 1.0\ \text{atm}$, $T = 300\ \text{K}$. From $pV = \nu RT$ and $\nu = m/M$: $p = \rho RT/M$, so: $$\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$$ Result: $\rho \approx 1.14\ \text{kg/m}^ Given Information See problem…
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Problem 2.3 — Barometric Formula: Pressure at Altitude
Problem Statement Solve the fluid mechanics problem: Find the air pressure at altitude $h = 5.0\ \text{km}$, assuming isothermal atmosphere at $T = 273\ \text{K}$ and sea-level pressure $p_0 = 1.013\times10^5\ \text{Pa}$. For an isothermal atmosphere, the pressure satisfies $dp = -\rho g\,dh$. With $\rho = pM/(RT)$: $$\frac{dp}{p} = -\frac{Mg}{RT}dh \ Given Information See problem…
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Problem 2.1 — Ideal Gas Law: Find Mass of Gas
Problem Statement Solve the thermodynamics problem: A vessel of volume $V = 2.0\ \text{L}$ contains an ideal gas at pressure $p = 3.0\ \text{atm}$ and temperature $T = 300\ \text{K}$. Find the mass of the gas if it is oxygen. The ideal gas law: $pV = \nu RT$, so $\nu = \frac{pV}{RT}$. Converting: $p =…
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Problem 2.2 — Isothermal Process: Pressure After Compression
Problem Statement Solve the thermodynamics problem: A vessel of volume $V_1 = 10\ \text{L}$ at $p_1 = 1.0\ \text{atm}$, $T = 300\ \text{K}$ is compressed isothermally to $V_2 = 4.0\ \text{L}$. Find the new pressure. Boyle’s Law at constant temperature: $p_1 V_1 = p_2 V_2$. $$p_2 = p_1 \frac{V_1}{V_2} = 1.0 \times \frac{10}{4.0} = 2.5\…