What is the initial absolute pressure?

Gauge pressure + atmospheric pressure = absolute pressure. Initial: $15 + 1 = 16$ atm. Final: $11 + 1 = 12$ atm.

How many moles initially?

$n_1 = \frac{P_1 V}{RT_1} = \frac{16 \times 101325 \times 0.03}{8.3 \times 300} = \frac{48636}{2490} \approx 19.5$ mol

How many moles finally?

$n_2 = \frac{P_2 V}{RT_2} = \frac{12 \times 101325 \times 0.03}{8.3 \times 290} = \frac{36477}{2407} \approx 15.2$ mol

What mass was withdrawn?

$\Delta n = n_1 - n_2 \approx 4.4$ mol. Mass $= 4.4 \times 32 \approx 141$ g

Why use absolute pressure in ideal gas law?

Ideal gas law $PV = nRT$ requires absolute pressure (in Pa or atm absolute). Gauge pressure = absolute - atmospheric. Always add 1 atm to convert gauge to absolute.

An oxygen cylinder of volume 30 L has an initial gauge pressure of 15 atm and temperature of $27°C$. After some oxygen i

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An oxygen cylinder of volume 30 L has an initial gauge pressure of 15 atm and temperature of $27°C$. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and temperature drops to $17°C$. What is the mass of oxygen taken out? (Atmospheric pressure $= 1$ atm, $R = 8.3$ J mol$^{-1}$K$^{-1}$, molar mass of O$_2 = 32$ g/mol)

Options

$141$ g

$0.141$ g

$0.44$ g

$44$ g

Correct Answer

141 g

Solution

Absolute pressures: $P_1 = 16$ atm, $P_2 = 12$ atm

Temperatures: $T_1 = 300$ K, $T_2 = 290$ K, Volume $V = 30$ L

$$n_1 = \frac{P_1 V}{RT_1} = \frac{16 \times 1 \times 30}{0.0821 \times 300} \approx 19.5 \text{ mol}$$

$$n_2 = \frac{P_2 V}{RT_2} = \frac{12 \times 1 \times 30}{0.0821 \times 290} \approx 15.1 \text{ mol}$$$$\Delta n = 19.5 - 15.1 = 4.4 \text{ mol}$$$$\text{Mass} = 4.4 \times 32 \approx \boxed{141 \text{ g}}$$

Gauge → Absolute: add 1 atm. Use $PV=nRT$. Withdrawn mass $= \Delta n \times M$

Theory: Kinetic Theory

1. Ideal Gas Law

$PV = nRT$ where $P$ = absolute pressure (Pa), $V$ = volume (m³), $n$ = moles, $R = 8.314$ J mol⁻¹K⁻¹, $T$ = temperature (K). Always use absolute pressure — gauge pressure must have atmospheric pressure (1 atm = 101325 Pa) added. $T$(K) $= T$(°C) $+ 273$. For same container with gas removed: volume constant, use $n_1$ and $n_2$ at respective $P$ and $T$.

2. Gauge vs Absolute Pressure

Absolute pressure = pressure measured from perfect vacuum. Gauge pressure = pressure above atmospheric (shown by most pressure gauges). Absolute = Gauge + Atmospheric. At sea level: $P_{atm} = 1$ atm = 101325 Pa = 760 mmHg = 1.013 bar. Tyre pressure 30 psi gauge = 44.7 psi absolute. Vacuum gauge: reads below atmospheric (negative gauge pressure). Blood pressure: systolic 120 mmHg gauge = 120 mmHg above atmospheric.

3. Molecular Speeds

RMS speed: $v_{rms} = \sqrt{3RT/M}$. Mean speed: $\bar{v} = \sqrt{8RT/\pi M}$. Most probable: $v_p = \sqrt{2RT/M}$. Ratio: $v_p : \bar{v} : v_{rms} = 1 : 1.128 : 1.225$. All $\propto \sqrt{T/M}$. Heavier molecules move slower. At same temperature, all gases have same average KE $= \frac{3}{2}k_BT$ per molecule.

4. Kinetic Theory Pressure

$P = \frac{1}{3}\rho v_{rms}^2 = \frac{1}{3}\frac{mN}{V}v_{rms}^2$. Pressure arises from molecular collisions with walls. Each collision transfers momentum $2mv$ to wall. Average force per unit area = pressure. This derivation assumes: elastic collisions, random motion, no intermolecular forces (ideal gas assumptions).

5. Degrees of Freedom

Each degree of freedom (DOF) contributes $\frac{1}{2}k_BT$ to average KE (equipartition). Monatomic (He, Ar): 3 translational DOF, $U = \frac{3}{2}nRT$, $C_v = \frac{3}{2}R$, $\gamma = 5/3$. Diatomic (N₂, O₂): 5 DOF (3 trans + 2 rot), $C_v = \frac{5}{2}R$, $\gamma = 7/5$. At very high T: vibrational DOF activated.

6. Van der Waals Gas

Real gas: $(P + a/V_m^2)(V_m - b) = RT$. $a$ = intermolecular attraction correction. $b$ = finite molecular volume. At high $P$ or low $T$: deviates from ideal. Compressibility factor $Z = PV/nRT$. Ideal: $Z=1$. Real: $Z < 1$ (moderate $P$, attraction dominant), $Z > 1$ (high $P$, repulsion dominant). Boyle temperature: $T_B = a/(Rb)$ where $Z \approx 1$.

7. Dalton Law of Partial Pressures

$P_{total} = P_1 + P_2 + \cdots$ (for ideal gas mixture). Partial pressure $P_i = x_i P_{total}$ where $x_i = n_i/n_{total}$ = mole fraction. Each gas behaves independently. Average molar mass of mixture: $M_{avg} = \sum x_i M_i$. Density of gas mixture: $\rho = P M_{avg}/(RT)$. Applications: air (N₂ + O₂ + Ar + CO₂ + ...), diving gas mixtures (adjust O₂ partial pressure for depth).

8. Mean Free Path

$\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$ where $d$ = molecular diameter, $n$ = number density. $\lambda \propto 1/P$ (inversely proportional to pressure). At 1 atm, 25°C: $\lambda_{air} \approx 68$ nm. At low pressure (10⁻³ atm): $\lambda \approx 68$ μm (comparable to small device sizes). Determines: viscosity, thermal conductivity, diffusion rate. Knudsen number $= \lambda/L$ determines if continuum mechanics applies.

Frequently Asked Questions

1. Why must gauge pressure be converted to absolute? ⌄

$PV = nRT$ is derived from kinetic theory where $P$ = force per unit area due to molecular collisions — this is absolute pressure. Gauge pressure (shown on most instruments) is excess above atmospheric. A tyre at "0 gauge pressure" (flat) still has 1 atm of gas inside! If we used gauge = 0, ideal gas law would give $n = 0$ (no gas) — clearly wrong. Always: Absolute = Gauge + Atmospheric (1 atm = 101325 Pa = 1.013×10⁵ Pa).

2. How is the calculation done step by step? ⌄

Step 1: Convert gauge to absolute: $P_1 = 15+1 = 16$ atm, $P_2 = 11+1 = 12$ atm. Step 2: Convert °C to K: $T_1 = 27+273 = 300$ K, $T_2 = 17+273 = 290$ K. Step 3: $n_1 = P_1V/RT_1 = (16\times30)/(0.0821\times300) = 480/24.63 \approx 19.49$ mol. Step 4: $n_2 = 12\times30/(0.0821\times290) = 360/23.81 \approx 15.12$ mol. Step 5: $\Delta n = 4.37$ mol. Step 6: Mass $= 4.37\times32 \approx 139.8 \approx 141$ g (small rounding differences give 141 g as closest option).

3. What is the value of R to use? ⌄

$R = 8.314$ J mol⁻¹K⁻¹ = 8.314 Pa·m³ mol⁻¹K⁻¹ (when $P$ in Pa, $V$ in m³). $R = 0.0821$ L·atm mol⁻¹K⁻¹ (when $P$ in atm, $V$ in litres). $R = 1.987$ cal mol⁻¹K⁻¹. For this problem with $P$ in atm and $V$ in litres: use $R = 0.0821$ L·atm mol⁻¹K⁻¹. If $P$ in Pa ($1$ atm = 101325 Pa) and $V$ in m³ ($30$ L $= 0.030$ m³): use $R = 8.314$.

4. Does temperature change affect the calculation significantly? ⌄

Yes. If temperature were constant ($T_1 = T_2 = 300$ K): $\Delta n = V/RT \times (P_1-P_2) = 30/(0.0821\times300)\times(16-12) = 1.217\times4 = 4.87$ mol → mass $= 155.8$ g. With temperature drop ($T_2=290$ K): $n_2$ increases slightly (same pressure but lower $T$ means more moles). This reduces $\Delta n$ slightly → $\approx 141$ g. The 10 K temperature drop matters — contributes about 10 g difference in the final answer.

5. What is the difference between gauge and absolute in daily life? ⌄

Car tyre: "32 psi" on tyre sidewall = 32 psi gauge = 32+14.7 = 46.7 psi absolute. Blood pressure: "120/80 mmHg" is gauge (above atmospheric). Actual absolute blood pressure = 120+760 = 880 mmHg systolic. Cooking pressure cooker: "15 psi" gauge cooking pressure = absolute 15+14.7 = 29.7 psi = about 2 atm absolute → water boils at ~121°C (higher boiling point at higher pressure). Weather barometer: reads absolute atmospheric pressure directly (760 mmHg = 1 atm at sea level).

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