Gas Laws and Kinetic Molecular Theory (4B)
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MCAT Chemical and Physical Foundations of Biological Systems › Gas Laws and Kinetic Molecular Theory (4B)
A physiological model treats inhaled air as an ideal gas. During inspiration, thoracic expansion increases lung volume while temperature and moles are approximately constant over a short interval. The question targets Boyle’s law.
What effect would an increase in lung volume most likely have on alveolar pressure (relative to atmospheric) during this interval?
A. Alveolar pressure increases above atmospheric
B. Alveolar pressure decreases below atmospheric
C. Alveolar pressure remains equal to atmospheric
D. Alveolar pressure becomes independent of volume because $R$ is constant
Alveolar pressure becomes independent of volume because $R$ is constant
Alveolar pressure remains equal to atmospheric
Alveolar pressure decreases below atmospheric
Alveolar pressure increases above atmospheric
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of Boyle's law applied to respiratory mechanics. During inspiration, thoracic expansion increases lung volume while temperature and moles remain approximately constant over short intervals. In this scenario, increased lung volume creates a pressure gradient according to Boyle's law (P₁V₁ = P₂V₂). The correct choice, B, follows because when lung volume increases, alveolar pressure decreases below atmospheric pressure, creating the pressure gradient that drives air flow into the lungs. Choice A is incorrect because it suggests pressure would increase with volume expansion, which would prevent inspiration. In similar questions, remember that breathing depends on pressure gradients created by volume changes.
A sample of an ideal gas is confined in a frictionless piston-cylinder assembly at constant temperature (isothermal). The gas is compressed quasi-statically from $2.0\ \mathrm{L}$ to $1.0\ \mathrm{L}$ with no change in moles of gas.
Based on Boyle’s law, what effect would this compression most likely have on the gas pressure?
The pressure would decrease because compression reduces the frequency of wall collisions.
The pressure would approximately double because $P\propto 1/V$ at constant $T$ and $n$.
The pressure would remain unchanged because temperature is constant and pressure depends only on $T$.
The pressure would approximately halve because $P\propto V$ at constant $T$ and $n$.
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of isothermal compression using Boyle's law. Boyle's law states that for a fixed amount of gas at constant temperature, pressure is inversely proportional to volume (P ∝ 1/V or PV = constant). In this scenario, the isothermal compression from 2.0 L to 1.0 L represents a halving of volume while maintaining constant T and n. The correct choice, A, follows because when volume is halved at constant T and n, pressure must double to maintain the constant PV product required by Boyle's law. Choice B is incorrect because it suggests direct proportionality between P and V, which contradicts the inverse relationship. In similar questions, remember that isothermal processes maintain constant temperature, and for ideal gases, this means PV remains constant throughout the process.
Two gases, He and O$_2$, are compared at the same temperature in a closed container. The analysis uses kinetic molecular theory for ideal gases.
Which prediction is most consistent with kinetic molecular theory at equal temperature?
A. He has a higher average translational kinetic energy than O$_2$
B. He and O$_2$ have the same average translational kinetic energy
C. O$_2$ has a higher average translational kinetic energy because it is heavier
D. The heavier gas must have a higher rms speed to maintain the same temperature
He has a higher average translational kinetic energy than O$_2$
The heavier gas must have a higher rms speed to maintain the same temperature
He and O$_2$ have the same average translational kinetic energy
O$_2$ has a higher average translational kinetic energy because it is heavier
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of average kinetic energy at equal temperature. Kinetic molecular theory states that average translational kinetic energy depends only on temperature: KE_avg = (3/2)kT, where k is Boltzmann's constant. In this scenario, He and O₂ are at the same temperature in the same container. The correct choice, B, follows because temperature is the sole determinant of average translational kinetic energy, regardless of molecular mass. Choice D is incorrect because it suggests heavier molecules have more kinetic energy at the same temperature, confusing kinetic energy with momentum. In similar questions, remember that while heavier molecules move slower at the same temperature, their average kinetic energy is identical.
In a respiratory physiology experiment, an alveolar gas sample is approximated as an ideal gas at constant temperature. The partial pressure of $\mathrm{O_2}$ is measured as $P_{\mathrm{O_2}}=100\ \mathrm{mmHg}$ in a mixture with total pressure $P_{\text{total}}=760\ \mathrm{mmHg}$. Assume the mixture behaves ideally.
Based on Dalton’s law of partial pressures, which value is most consistent with the mole fraction of $\mathrm{O_2}$ in the sample?
$x_{\mathrm{O_2}}\approx 7.6$
$x_{\mathrm{O_2}}\approx 0.76$
$x_{\mathrm{O_2}}\approx 0.13$
$x_{\mathrm{O_2}}\approx 0.013$
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of Dalton's law and mole fractions. Dalton's law states that in a gas mixture, each component's partial pressure equals its mole fraction times total pressure: Pi = xiPtotal, which rearranges to xi = Pi/Ptotal. In this scenario, oxygen has partial pressure 100 mmHg in a mixture with total pressure 760 mmHg. The correct choice, A, follows because xO₂ = PO₂/Ptotal = 100/760 ≈ 0.13, representing oxygen as about 13% of the gas mixture by moles. Choice C is incorrect because it appears to calculate 760-100 = 660 then divide by total pressure, which has no physical meaning for mole fraction. In similar questions, remember that mole fractions must sum to 1.0 for all components and individual mole fractions must be between 0 and 1.
A gas-phase reaction is monitored in a closed, rigid reactor at constant temperature. Initially, the reactor contains $1.0\ \mathrm{mol}$ of an ideal gas at $P_0$. A valve then opens to a second, identical evacuated rigid chamber, allowing the gas to expand freely until equilibrium. Temperature remains constant and no gas is lost.
Based on the ideal gas law applied to free expansion into a vacuum, what is the most consistent prediction for the final pressure in the two-chamber system?
The final pressure is $P_0$ because temperature and moles are unchanged.
The final pressure is $\tfrac{1}{2}P_0$ because total volume doubles at constant $T$ and $n$.
The final pressure cannot be predicted without the gas molar mass.
The final pressure is $2P_0$ because the gas occupies twice the space and collides more often.
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of free expansion into vacuum. When an ideal gas expands into vacuum at constant temperature, the process is isothermal with no work done, and the ideal gas law applies to the final equilibrium state. In this scenario, the gas initially in volume V expands to fill total volume 2V while maintaining constant T and n. The correct choice, C, follows because from PV = nRT, if volume doubles while T and n remain constant, pressure must halve to maintain the equality, giving Pfinal = P₀/2. Choice A is incorrect because it suggests pressure increases with volume, contradicting the inverse relationship at constant T and n. In similar questions involving free expansion, apply the ideal gas law to initial and final states, recognizing that doubling volume halves pressure when other variables are constant.
A closed container of fixed volume holds an ideal gas. The researcher increases the number of moles of gas by injecting additional gas while maintaining constant temperature (thermal bath at 298 K). Principle: at constant $T$ and $V$, the ideal gas law implies $P \propto n$. If the initial pressure is $P_1=1.0\ \text{atm}$ at $n_1=0.50\ \text{mol}$, what is the most consistent prediction for the pressure after increasing to $n_2=0.80\ \text{mol}$?
Constants: none needed.
$P_2=1.3\ \text{atm}$ because pressure increases by $0.3\ \text{atm}$ when moles increase by 0.30 mol
$P_2=0.625\ \text{atm}$ because pressure is inversely proportional to moles at constant temperature
$P_2=1.6\ \text{atm}$ because $P_2=P_1,(n_2/n_1)$ at constant temperature and volume
$P_2=2.56\ \text{atm}$ because pressure scales with $n^2$ for ideal gases at constant volume
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of pressure changes with varying amounts of gas. The ideal gas law shows that at constant temperature and volume, pressure is directly proportional to the number of moles (P ∝ n). In this scenario, moles increase from 0.50 mol to 0.80 mol while temperature (298 K) and volume remain constant. The correct choice, B, follows because P₂ = P₁(n₂/n₁) = 1.0 atm × (0.80 mol/0.50 mol) = 1.0 atm × 1.6 = 1.6 atm. Choice A is incorrect because it suggests inverse proportionality between pressure and moles, contradicting the ideal gas law. In similar questions, ensure temperature and volume are constant, then apply direct proportionality between pressure and amount of gas.
A gas sample is confined under a frictionless piston at constant external pressure $P=1.00\ \text{atm}$, allowing volume to change as temperature changes. A researcher measures $V$ at different $T$ for $n$ fixed. Data: $T$ (K) → $V$ (L): 250→6.2; 275→6.8; 300→7.4; 325→8.0; 350→8.6. (Constants: $R=0.0821\ \text{L·atm·mol}^{-1}\text{·K}^{-1}$.) Using Charles’s law ($V\propto T$ at constant $P,n$), which prediction aligns with the observed behavior if the temperature is increased from 300 K to 360 K?
The volume should remain near 7.4 L because the pressure is fixed at 1 atm.
The volume should decrease because higher temperature increases pressure, compressing the gas.
The volume should increase proportionally, to about $V\approx 7.4\times(360/300)\approx 8.9\ \text{L}$.
The volume should increase, but only if the number of moles increases.
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of the volume-temperature relationship at constant pressure. Gas laws, such as Charles’s law, describe the relationship between volume and absolute temperature under constant pressure and moles. In this scenario, the frictionless piston allowing volume changes at fixed pressure illustrates the direct proportionality between volume and temperature. The correct choice, B, follows because extrapolating the data predicts V ≈ 7.4 × (360/300) ≈ 8.9 L, consistent with V ∝ T. Choice A is incorrect because it misapplies Gay-Lussac’s law, ignoring that volume can change here. In similar questions, ensure to check if pressure and moles are constant to apply Charles’s law accurately. Verify calculations using absolute temperatures to avoid common errors with Celsius.
A sample of gas is held at constant temperature in a cylinder with a frictionless piston. The investigator increases the external pressure in steps, allowing the system to equilibrate each time, and records the corresponding gas volume. The product $PV$ is approximately constant across the measurements. Which conclusion about gas behavior is most consistent with Boyle’s law under these conditions? (Use $R = 0.0821\ \text{L·atm·mol}^{-1}\text{·K}^{-1}$.)
Volume is directly proportional to the number of moles, so changing pressure cannot change volume at fixed $n$.
Volume decreases as pressure increases such that $PV$ remains approximately constant at fixed temperature.
Pressure is inversely proportional to temperature, so keeping temperature constant forces $P$ to remain constant.
Volume increases as pressure increases because $PV$ must increase when pressure increases.
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of Boyle's law experimental verification. Boyle's law states that at constant temperature, pressure and volume are inversely proportional, meaning their product PV remains constant. In this scenario, the cylinder with frictionless piston allows volume to adjust as external pressure changes while maintaining constant temperature. The correct choice, A, follows because the data show PV remaining approximately constant, confirming that volume decreases as pressure increases in an inverse relationship (P₁V₁ = P₂V₂). Choice B is incorrect because it suggests volume increases with pressure, which would violate the inverse relationship fundamental to Boyle's law. In similar questions, verify that the product PV remains constant across measurements to confirm Boyle's law behavior.
In a respiratory physiology experiment, an alveolar-sized chamber is modeled as a fixed-volume compartment (volume constant over the measurement interval). The gas mixture is rapidly warmed from 310 K to 330 K without changing the number of moles present. According to kinetic molecular theory and the ideal-gas relationship, which prediction aligns with the expected change in pressure in the compartment? (Use $R = 0.0821\ \text{L·atm·mol}^{-1}\text{·K}^{-1}$.)
Pressure increases only if the gas is non-ideal; an ideal gas would show no change with temperature.
Pressure increases because higher temperature increases average translational kinetic energy and collision impulse at the walls.
Pressure decreases because higher temperature reduces the collision frequency at fixed volume.
Pressure is unchanged because kinetic energy changes do not affect macroscopic pressure at fixed volume.
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of pressure changes in a fixed-volume respiratory model. According to kinetic molecular theory, temperature is directly related to the average kinetic energy of gas molecules, and pressure results from molecular collisions with container walls. In this scenario, the alveolar chamber has fixed volume while temperature increases from 310 K to 330 K. The correct choice, A, follows because higher temperature increases average molecular kinetic energy, leading to more forceful collisions with walls and thus higher pressure, consistent with Gay-Lussac's law (P ∝ T at constant V). Choice B is incorrect because it claims pressure decreases with temperature, contradicting both kinetic theory and Gay-Lussac's law. In similar questions, remember that at constant volume, pressure increases linearly with absolute temperature due to increased molecular kinetic energy.
To isolate the effect of volume on pressure, 0.0500 mol of He(g) is maintained at 298 K in a cylinder with a movable piston. The gas is compressed quasi-statically from 4.00 L to 2.00 L while temperature is held constant by a thermal reservoir. Which prediction based on Boyle’s law ($P \propto 1/V$ at constant $T,n$) is most consistent with the manipulation? (Use $R = 0.0821\ \text{L·atm·mol}^{-1}\text{·K}^{-1}$.)
The final pressure will be unchanged because temperature is held constant.
The final pressure will increase by a factor of 4 because pressure scales with $1/V^2$ at constant $T$.
The final pressure will be half the initial pressure because volume decreased by a factor of 2.
The final pressure will be approximately twice the initial pressure because volume decreased by a factor of 2.
Explanation
This question assesses understanding of gas laws and kinetic molecular theory (4B) in the context of Boyle's law during isothermal compression. Boyle's law states that pressure is inversely proportional to volume (P ∝ 1/V) when temperature and amount of gas are held constant. In this scenario, helium gas is compressed from 4.00 L to 2.00 L at constant temperature, representing a volume decrease by a factor of 2. The correct choice, B, follows because when volume is halved at constant temperature, pressure must double according to P₁V₁ = P₂V₂. Choice A is incorrect because it suggests pressure decreases when volume decreases, which contradicts the inverse relationship in Boyle's law. In similar questions, ensure temperature remains constant throughout the process and apply the relationship P₁V₁ = P₂V₂ to find the final pressure.