Understand Gas Laws - IB Physics

The microscopic origin of the pressure exerted by a gas on the walls of its container is the
Answer: rate of change of momentum of molecules colliding with the wall. According to the kinetic theory of gases, pressure is a result of the constant collisions of gas molecules with the container walls. Each collision involves a change in the molecule's momentum. By Newton's second law, a change in momentum implies a force exerted on the molecule by the wall, and by Newton's third law, the molecule exerts an equal and opposite force on the wall. The macroscopic pressure is the average force per unit area from these countless collisions, directly related to the rate of change of momentum.
A gas undergoes a cyclic process consisting of three steps: (1) isobaric expansion from volume $V$ to $2V$ at pressure $P$ , (2) isochoric cooling back to the original temperature, and (3) isothermal compression back to the original state. What is the pressure after step 2?
Answer: The pressure after step 2 is $\frac{P}{2}$ When analyzing cyclic thermodynamic processes, you need to track how pressure, volume, and temperature change through each step using the ideal gas law: $PV = nRT$ . Let's trace through each step systematically. Initially, we have state ( $P$ , $V$ , $T_0$ ). In step 1, isobaric expansion doubles the volume at constant pressure $P$ . Since $PV = nRT$ , when volume doubles at constant pressure, temperature must also double to $2T_0$ . So after step 1: ( $P$ , $2V$ , $2T_0$ ). Step 2 is isochoric cooling "back to the original temperature" $T_0$ at constant volume $2V$ . Using the ideal gas law again: if temperature halves while volume stays constant, pressure must also halve. Therefore: $P_2 = \frac{P}{2}$ at state ( $\frac{P}{2}$ , $2V$ , $T_0$ ). Now let's examine the wrong answers. Answer A ( $\frac{P}{4}$ ) would occur if you incorrectly thought both volume and temperature changes affected pressure multiplicatively. Answer B ( $P$ ) assumes pressure somehow returns to its original value, ignoring that we're at a different volume-temperature combination. Answer C ( $2P$ ) incorrectly suggests pressure increases during cooling at constant volume. The key insight is that answer D ( $\frac{P}{2}$ ) correctly applies the ideal gas law: when temperature decreases by half at constant volume, pressure must decrease proportionally. Study tip: For thermodynamic cycles, always track all three variables (P, V, T) through each step using $PV = nRT$ . Draw a table or diagram to avoid losing track of the state changes.
An ideal gas is compressed at a constant temperature. Which statement correctly describes the changes to the gas molecules?
Answer: Their average kinetic energy is unchanged, and they collide with the walls more frequently. Constant temperature implies that the average kinetic energy of the gas molecules ( $\frac{3}{2}k_B T$ ) remains unchanged, and thus their average speed is also unchanged. Compression means the volume decreases. With the same speed in a smaller volume, molecules will travel shorter distances before hitting a wall, so they collide with the container walls more frequently. This increased collision rate is what causes the pressure to rise (Boyle's Law).
A fixed mass of an ideal gas undergoes a process where its pressure is tripled and its volume is reduced to one-third of its initial value. What is the effect on its root-mean-square (RMS) molecular speed?
Answer: It remains unchanged. The RMS speed $v_{rms}$ is proportional to the square root of the absolute temperature $T$ . We must first determine the change in temperature using the combined gas law: $T_2/T_1 = (P_2/P_1)(V_2/V_1)$ . We are given $P_2 = 3P_1$ and $V_2 = V_1/3$ . Therefore, $T_2/T_1 = (3)(1/3) = 1$ . The temperature does not change. Since $v_{rms}$ depends only on temperature for a given gas, the RMS speed remains unchanged.
Container X has volume V and holds an ideal gas at pressure P. Container Y has volume 2V and holds the same ideal gas, also at pressure P. Both containers are at the same temperature. A valve connecting them is opened, and the gas reaches equilibrium while the temperature is kept constant. What is the final pressure?
Answer: 2P/3 Initially, container X has nX = PV/(RT) moles and container Y has nY = P(2V)/(RT) moles. The total number of moles is ntotal = PV/(RT) + 2PV/(RT) = 3PV/(RT). After opening the valve, the total volume is V + 2V = 3V. Using PV = nRT for the final state: Pfinal = ntotal RT/(3V) = [3PV/(RT)] × RT/(3V) = P × (3V/3V) = 2P/3.
A container of volume 10.0 m³ contains 2.0 moles of a monatomic ideal gas. The total internal energy of the gas is 7.50 × 10³ J. What is the pressure of the gas? (Ideal gas constant R = 8.31 J mol⁻¹ K⁻¹)
Answer: 500 Pa For a monatomic ideal gas, the internal energy is U = (3/2)nRT. From the ideal gas law PV = nRT, we can derive that nRT = PV. Substituting this into the internal energy equation gives U = (3/2)PV. Rearranging for pressure: P = 2U/(3V) = (2 × 7.50 × 10³ J)/(3 × 10.0 m³) = 15000/30 = 500 Pa.
An ideal gas in a cylinder with a movable piston undergoes the process shown in the P-V diagram. The gas starts at point A ( $2.0 \text{ atm}$ , $1.0 \text{ L}$ ), undergoes isobaric expansion to point B ( $2.0 \text{ atm}$ , $3.0 \text{ L}$ ), then isochoric cooling to point C ( $P_C$ , $3.0 \text{ L}$ ). If the temperature at point A is $300 \text{ K}$ , what is the pressure at point C?
Answer: The pressure at point C is $0.67 \text{ atm}$ First find temperature at B using Charles's law (isobaric process): $\frac{V_A}{T_A} = \frac{V_B}{T_B}$ , so $T_B = T_A \times \frac{V_B}{V_A} = 300 \times \frac{3.0}{1.0} = 900\text{ K}$ . For isochoric process B→C, the gas returns to original temperature $T_C = T_A = 300\text{ K}$ . Using Gay-Lussac's law: $\frac{P_B}{T_B} = \frac{P_C}{T_C}$ , so $P_C = P_B \times \frac{T_C}{T_B} = 2.0 \times \frac{300}{900} = 0.67\text{ atm}$ .
A vertical cylinder contains a gas separated from the atmosphere by a movable piston of mass $m$ and cross-sectional area $A$ . The gas initially has volume $V_0$ at temperature $T_0$ . When the temperature is increased to $2T_0$ , the piston moves up by distance $h$ . What is the relationship between the initial gas pressure and atmospheric pressure $P_{atm}$ ?
Answer: $P_0 = P_{atm} + \frac{mg}{A}$ and $h = \frac{V_0}{A}$ When you encounter a gas-in-cylinder problem with a movable piston, you need to analyze two key aspects: the force equilibrium on the piston and how the gas volume changes with temperature. For the gas pressure, consider the forces acting on the piston. The gas pushes up with force $P_0 A$ , while atmospheric pressure pushes down with force $P_{atm} A$ and gravity pulls down with force $mg$ . Since the piston is in equilibrium: $P_0 A = P_{atm} A + mg$ , which gives $P_0 = P_{atm} + \frac{mg}{A}$ . For the volume change, use Charles's Law since pressure remains constant (the piston can move freely). When temperature doubles from $T_0$ to $2T_0$ , the volume doubles from $V_0$ to $2V_0$ . The volume increase is $\Delta V = 2V_0 - V_0 = V_0$ . Since this volume change equals the cylinder's cross-sectional area times the height increase: $V_0 = A \times h$ , so $h = \frac{V_0}{A}$ . Choice A incorrectly states $h = \frac{2V_0}{A}$ , which would mean the gas volume tripled rather than doubled. Choice B has the wrong pressure relationship ( $P_0 = P_{atm} - \frac{mg}{A}$ ), suggesting the gas pressure is less than atmospheric pressure, which is impossible since the piston's weight adds to the downward force. Choice D ignores the piston's weight entirely, setting $P_0 = P_{atm}$ . Remember: in piston problems, always account for the piston's weight in your force analysis, and use the appropriate gas law based on what remains constant during the process.
A balloon contains helium gas at $15°\text{C}$ and $1.0 \text{ atm}$ . The balloon is taken to an altitude where the external pressure drops to $0.5 \text{ atm}$ and the temperature drops to $-25°\text{C}$ . Assuming the balloon material is perfectly flexible and the amount of gas remains constant, by what factor does the volume change?
Answer: The volume increases by a factor of $1.74$ When you encounter gas problems involving changes in pressure, temperature, and volume, you're dealing with the combined gas law. Since the balloon is perfectly flexible and the gas amount remains constant, you can use $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ to find how volume changes. First, convert temperatures to Kelvin: $T_1 = 15°C + 273 = 288K$ and $T_2 = -25°C + 273 = 248K$ . The initial conditions are $P_1 = 1.0 \text{ atm}$ , and final conditions are $P_2 = 0.5 \text{ atm}$ . Rearranging for the volume ratio: $\frac{V_2}{V_1} = \frac{P_1T_2}{P_2T_1} = \frac{(1.0)(248)}{(0.5)(288)} = \frac{248}{144} = 1.72$ This rounds to $1.74$ , confirming answer D is correct. Answer A ( $1.44$ ) likely comes from forgetting to convert Celsius to Kelvin and using the temperatures directly in the calculation. Answer B ( $2.00$ ) represents the pressure ratio alone ( $\frac{1.0}{0.5} = 2$ ), ignoring the temperature decrease that partially counteracts the volume expansion. Answer C ( $0.87$ ) appears to invert the pressure ratio, perhaps calculating $\frac{P_2}{P_1} \times \frac{T_2}{T_1}$ instead of the correct relationship. Remember that gas law problems always require absolute temperature (Kelvin), and when multiple variables change simultaneously, you must account for all of them. The pressure decrease tends to expand the gas while the temperature decrease tends to compress it—the net effect determines the final volume change.
An ideal gas has a molar mass $M$ . Which expression represents the density $\rho$ of the gas in terms of pressure $P$ , absolute temperature $T$ , and the ideal gas constant $R$ ?
Answer: $\frac{PM}{RT}$ Start with the ideal gas law: $PV = nRT$ . The number of moles $n$ is the total mass $m$ divided by the molar mass $M$ , so $n = m/M$ . Substituting this gives $PV = (m/M)RT$ . Density $\rho$ is mass per unit volume, $\rho = m/V$ . Rearranging the equation to solve for $m/V$ gives $m/V = PM/RT$ . Therefore, $\rho = PM/RT$ .
A sealed cubic container with side length $L$ contains an ideal gas at pressure $P$ . What is the average magnitude of the force exerted by the gas on one face of the container?
Answer: $P L^2$ Pressure is defined as force per unit area, $P = F/A$ . Therefore, the force exerted is $F = P \times A$ . The area of one face of a cube with side length $L$ is $A = L^2$ . Thus, the average force on one face is $F = P L^2$ . $L^3$ is the volume, and $6L^2$ is the total surface area of all six faces.
A bubble of an ideal gas rises from the bottom of a lake to the surface. Its volume increases by a factor of three. The water temperature is constant throughout the lake. What is the ratio of the pressure at the bottom of the lake to the pressure at the surface?
Answer: 3 Since the temperature is constant and the mass of gas in the bubble is constant, Boyle's Law applies: $P_1V_1 = P_2V_2$ . Let state 1 be at the bottom and state 2 be at the surface. We are given $V_2 = 3V_1$ . We need to find the ratio $P_1/P_2$ . From Boyle's Law, $P_1/P_2 = V_2/V_1$ . Substituting the given information, $P_1/P_2 = (3V_1)/V_1 = 3$ .
A sealed container holds an ideal gas at temperature $T_1 = 300 \text{ K}$ and pressure $P_1 = 2.0 \times 10^5 \text{ Pa}$ . The gas is heated until its pressure doubles, then expanded isothermally until its pressure returns to the original value. What is the final volume of the gas compared to its initial volume?
Answer: The final volume is twice the initial volume Using the ideal gas law for each step: Initially $PV = nRT_1$ . After heating at constant volume: $P_2 = 2P_1$ so $T_2 = 2T_1 = 600\text{ K}$ . During isothermal expansion: $P_2V_2 = P_1V_f$ where $V_f$ is final volume. Since $P_2 = 2P_1$ : $2P_1V_2 = P_1V_f$ , so $V_f = 2V_2 = 2V_1$ . The final volume is twice the initial volume.
According to the kinetic theory of gases, which assumption is made for an ideal gas?
Answer: The volume of the molecules is negligible compared to the container volume. The kinetic theory of ideal gases is based on several assumptions. Key assumptions include: collisions between molecules (and with the walls) are perfectly elastic; there are no intermolecular forces between molecules; and the volume occupied by the molecules themselves is negligible compared to the volume of the container. Molecules move with a range of speeds (a distribution), not all with the same speed.
The pressure of a fixed mass of an ideal gas is doubled, and its volume is halved. What is the ratio of the final absolute temperature to the initial absolute temperature?
Answer: 1 Using the combined gas law, $P_1V_1/T_1 = P_2V_2/T_2$ . We are given $P_2 = 2P_1$ and $V_2 = V_1/2$ . We want to find the ratio $T_2/T_1$ . Rearranging the equation gives $T_2/T_1 = (P_2/P_1)(V_2/V_1)$ . Substituting the given values: $T_2/T_1 = (2P_1/P_1)( (V_1/2) / V_1) = (2)(1/2) = 1$ . The final temperature is the same as the initial temperature.
The pressure $P$ of an ideal gas is given by the kinetic theory equation $P = \frac{1}{3}\rho \langle c^2 \rangle$ , where $\rho$ is the gas density and $\langle c^2 \rangle$ is the mean-square speed of the molecules. A fixed mass of ideal gas in a rigid container is heated from 300 K to 900 K. By what factor does the term $\rho \langle c^2 \rangle$ increase?
Answer: 3 For a fixed mass of gas in a rigid container, both the number of moles $n$ and the volume $V$ are constant. From the ideal gas law $PV=nRT$ , pressure is directly proportional to absolute temperature, $P \propto T$ . If the temperature triples (from 300 K to 900 K), the pressure also triples. The kinetic theory equation states $P = \frac{1}{3}\rho \langle c^2 \rangle$ , which can be written as $P \propto \rho \langle c^2 \rangle$ . Since the pressure triples, the term $\rho \langle c^2 \rangle$ must also increase by a factor of 3.
A sealed container of fixed volume contains a mixture of two ideal gases: $2.0 \text{ mol}$ of gas A and $3.0 \text{ mol}$ of gas B. The total pressure is $5.0 \text{ atm}$ at $25°\text{C}$ . If the container is heated until the partial pressure of gas A becomes $4.0 \text{ atm}$ , what is the final temperature?
Answer: The final temperature is $596\text{ K}$ or $323°\text{C}$ When you encounter gas mixture problems involving temperature changes, you need to apply Dalton's Law of Partial Pressures combined with Gay-Lussac's Law. The key insight is that partial pressure is proportional to both the number of moles and temperature. First, find gas A's initial partial pressure. Since partial pressure depends on mole fraction: $P_A = \frac{n_A}{n_{total}} \times P_{total} = \frac{2.0}{5.0} \times 5.0 = 2.0 \text{ atm}$ For a fixed volume and amount of gas, partial pressure is directly proportional to absolute temperature: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ Converting the initial temperature: $T_1 = 25°C + 273 = 298 K$ Solving for final temperature: $T_2 = T_1 \times \frac{P_2}{P_1} = 298 K \times \frac{4.0}{2.0} = 596 K$ Converting back: $596 - 273 = 323°C$ Answer C ( $596 K$ or $323°C$ ) is correct. Answer A ( $477 K$ ) incorrectly assumes some other pressure ratio. Answer B ( $372 K$ ) might result from using the wrong initial partial pressure or incorrect temperature conversion. Answer D ( $298 K$ ) represents no temperature change, ignoring that pressure doubled. Remember: In gas mixture problems, always calculate partial pressures using mole fractions first, then apply gas laws to individual components. The partial pressure of one gas changing tells you about the overall temperature change affecting the entire mixture.
Two different monatomic ideal gases, X and Y, are mixed in a container and are in thermal equilibrium. The mass of a molecule of gas Y is four times the mass of a molecule of gas X ( $m_Y = 4m_X$ ). What is the ratio of the root-mean-square (RMS) speed of molecules of X to that of molecules of Y ( $v_{rms,X} / v_{rms,Y}$ )?
Answer: 2 Since the gases are in thermal equilibrium, they are at the same temperature. The average kinetic energy of the molecules of both gases is the same ( $E_k = \frac{3}{2}k_B T$ ). The kinetic energy is given by $\frac{1}{2}mv_{rms}^2$ . Therefore, $\frac{1}{2}m_X v_{rms,X}^2 = \frac{1}{2}m_Y v_{rms,Y}^2$ . This simplifies to $v_{rms,X}^2 / v_{rms,Y}^2 = m_Y / m_X$ . Given $m_Y = 4m_X$ , the ratio is 4. Taking the square root gives $v_{rms,X} / v_{rms,Y} = \sqrt{4} = 2$ .
Under which of the following conditions does a real gas most closely approximate the behavior of an ideal gas?
Answer: Low pressure and high temperature. An ideal gas is modeled as having particles with negligible volume and no intermolecular forces. Real gases approximate this behavior at low pressure, where the molecules are far apart, making their individual volumes negligible compared to the container volume. They also approximate it at high temperature, where the molecules' high kinetic energy is much greater than the potential energy of any intermolecular forces, making those forces insignificant.
A sealed, rigid container holds a fixed mass of an ideal gas. The absolute temperature of the gas is doubled. What is the effect on the pressure of the gas and the average translational kinetic energy of a gas molecule?
Answer: The pressure is doubled, and the average kinetic energy is doubled. For an ideal gas at constant volume, pressure is directly proportional to absolute temperature (from PV = nRT, since V and n are constant). Therefore, doubling the temperature doubles the pressure. The average translational kinetic energy of gas molecules is given by (3/2)kT, so it is also directly proportional to absolute temperature. Doubling the temperature therefore doubles the average kinetic energy.

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