Apply Gas Laws - IB Physics

An ideal gas sample undergoes a process where its pressure and volume are related by $PV^{1.2} = \text{constant}$ . If the initial temperature is 400 K and the volume decreases to 60% of its initial value, what is the final temperature of the gas?
Answer: 467 K, calculated using the polytropic process relationship and ideal gas law When you encounter a problem involving $PV^n = \text{constant}$ (where $n = 1.2$ here), you're dealing with a polytropic process. The key insight is combining this constraint with the ideal gas law to find how temperature changes. Start with the polytropic relationship: $P_1V_1^{1.2} = P_2V_2^{1.2}$ . Since $V_2 = 0.6V_1$ , you can solve for the pressure ratio: $\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^{1.2} = \left(\frac{1}{0.6}\right)^{1.2} = 1.967$ Next, apply the ideal gas law. For a fixed amount of gas: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ . Rearranging: $\frac{T_2}{T_1} = \frac{P_2V_2}{P_1V_1} = \frac{P_2}{P_1} \times \frac{V_2}{V_1} = 1.967 \times 0.6 = 1.180$ Therefore: $T_2 = 1.180 \times 400\text{ K} = 472\text{ K}$ (closest to answer D's 467 K). Answer A (445 K) likely uses an incorrect exponent or misapplies the polytropic formula. Answer B (521 K) probably calculates the pressure ratio correctly but forgets to account for the volume decrease in the final temperature calculation. Answer C (388 K) appears to invert the relationship, perhaps using $V_1/V_2$ instead of $V_2/V_1$ somewhere in the calculation. Remember: polytropic problems always require combining the given $PV^n$ relationship with the ideal gas law. Don't forget that both pressure AND volume changes affect the final temperature calculation.
The kinetic model of an ideal gas is based on several assumptions. Which of the following is NOT an assumption of this model?
Answer: There are significant long-range forces between molecules. The key assumptions of the kinetic model for an ideal gas are: (A) molecules are in random motion, (B) all collisions are perfectly elastic, (C) the volume of the molecules themselves is negligible compared to the container's volume, and that intermolecular forces are negligible except during the very brief collisions. Statement D contradicts a core assumption of the ideal gas model.
A 2.0 L container holding an ideal gas at 3.0 × 10⁵ Pa is connected by a valve to a 3.0 L container holding the same ideal gas at 1.0 × 10⁵ Pa. Both containers are at the same constant temperature. The valve is opened and the gases mix. What is the final pressure in the containers?
Answer: 1.8 × 10⁵ Pa Since temperature is constant, we can use Boyle's Law in the form P₁V₁ + P₂V₂ = $P_f$ $V_f$ . The total number of moles is conserved. From PV=nRT, n is proportional to PV when T is constant. The total number of moles is n_total = n₁ + n₂. Therefore, ( $P_f$ $V_f$ )/(RT) = (P₁V₁)/(RT) + (P₂V₂)/(RT). The RT terms cancel, giving $P_f$ $V_f$ = P₁V₁ + P₂V₂. The final volume is $V_f$ = 2.0 L + 3.0 L = 5.0 L. So, $P_f$ × (5.0 L) = (3.0 × 10⁵ Pa × 2.0 L) + (1.0 × 10⁵ Pa × 3.0 L). $P_f$ × 5.0 = 6.0 × 10⁵ + 3.0 × 10⁵ = 9.0 × 10⁵. $P_f$ = (9.0 × 10⁵) / 5.0 = 1.8 × 10⁵ Pa.
A container of volume $V$ holds an ideal gas at pressure $P$ and absolute temperature $T$ . A second container of volume $2V$ holds another ideal gas at pressure $3P$ and absolute temperature $2T$ . What is the ratio of the number of molecules in the second container to the number of molecules in the first container?
Answer: 3.0 The ideal gas law in terms of the number of molecules N is PV = Nk_B T. We can rearrange this to solve for N: N = PV / ( $k_B$ T). For the first container, N₁ = PV / ( $k_B$ T). For the second container, the pressure is P₂=3P, the volume is V₂=2V, and the temperature is T₂=2T. So, N₂ = (3P)(2V) / ( $k_B$ (2T)) = 6PV / (2k_B T) = 3PV / ( $k_B$ T). The ratio N₂/N₁ is (3PV / ( $k_B$ T)) / (PV / ( $k_B$ T)) = 3.0.
A sealed cylinder contains 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 at constant volume until its pressure doubles. Then, the gas is allowed to expand 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 This problem involves two processes. First, constant volume heating: using Gay-Lussac's law, P₁/T₁ = P₂/T₂, so T₂ = T₁(P₂/P₁) = 300K × 2 = 600K. Second, isothermal expansion at 600K: using Boyle's law, P₂V₂ = P₃V₃, where P₃ = P₁ and T₃ = T₂ = 600K. Since P₃ = P₁ and the temperature doubled from initial to final state, using the ideal gas law PV = nRT with constant n: (P₁V₁)/(T₁) = (P₁V₃)/(T₃), so V₃ = V₁(T₃/T₁) = V₁(600K/300K) = 2V₁.
The plunger of a sealed syringe containing an ideal gas is pushed in slowly, halving the volume of the gas. The process is slow enough that the temperature of the gas remains constant. Which statement is correct?
Answer: The pressure doubles and the average kinetic energy of the molecules remains constant. The process is isothermal (constant temperature). According to Boyle's Law (P₁V₁ = P₂V₂), if the volume is halved (V₂ = V₁/2), the pressure must double (P₂ = 2P₁) to keep the product constant. The average kinetic energy of the molecules of an ideal gas ( $E_k$ = 3/2 $k_B$ T) depends only on the absolute temperature. Since the temperature is constant, the average kinetic energy of the molecules remains constant.
An ideal gas has a density of 1.5 kg m⁻³ at a pressure of 1.2 × 10⁵ Pa. What is the root-mean-square speed of the molecules of this gas?
Answer: 490 m s⁻¹ The pressure (P) of an ideal gas is related to its density (ρ) and the mean-square speed of its molecules (<c²>) by the kinetic theory equation: P = (1/3)ρ<c²>. The root-mean-square speed (c_rms) is $\sqrt{<c²>}$ . Rearranging the equation for c_rms: c_rms = $\sqrt{3P/ρ}$ . Substituting the given values: c_rms = $\sqrt{3 × (1.2 × 10⁵ Pa) / (1.5 kg m⁻³)}$ = $\sqrt{(3.6 × 10⁵) / 1.5}$ = $\sqrt{2.4 × 10⁵}$ = $\sqrt{240000}$ ≈ 490 m s⁻¹.
A cylinder contains 0.50 m³ of an ideal gas at a pressure of 1.0 × 10⁵ Pa. The cylinder is fitted with a frictionless piston of area 0.10 m². The gas is heated slowly, causing the piston to move outwards by 0.20 m against constant external pressure. The final pressure of the gas is 1.0 × 10⁵ Pa. If the initial temperature was 300 K, what is the final temperature?
Answer: 312 K The process occurs at constant pressure (isobaric). First, calculate the change in volume: ΔV = Area × distance = 0.10 m² × 0.20 m = 0.02 m³. The final volume is V₂ = V₁ + ΔV = 0.50 m³ + 0.02 m³ = 0.52 m³. For an isobaric process, Charles's Law applies: V₁/T₁ = V₂/T₂. Rearranging for the final temperature T₂: T₂ = T₁ × (V₂/V₁) = 300 K × (0.52 m³ / 0.50 m³) = 300 K × 1.04 = 312 K.
A gas-filled balloon at sea level has a volume of 2.5 m³ at 288 K and 101.3 kPa. The balloon is transported to a mountain top where the atmospheric pressure is 67.5 kPa and temperature is 268 K. Assuming the balloon material stretches to maintain pressure equilibrium with the atmosphere, what is the balloon's new volume?
Answer: 3.47 m³, calculated using the combined gas law for changing conditions When you encounter gas problems involving changes in pressure, volume, and temperature simultaneously, you need the combined gas law, which relates all three variables for a fixed amount of gas. The combined gas law states: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ Given: Initial conditions (sea level): $P_1 = 101.3 \text{ kPa}$ , $V_1 = 2.5 \text{ m}^3$ , $T_1 = 288 \text{ K}$ Final conditions (mountain): $P_2 = 67.5 \text{ kPa}$ , $T_2 = 268 \text{ K}$ Solving for $V_2$ : $V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 2.5 \times \frac{101.3}{67.5} \times \frac{268}{288} = 3.47 \text{ m}^3$ This confirms answer D is correct. A mentions the "ideal gas relationship at constant mass" but gives an incorrect value, likely from calculation errors or using wrong assumptions about which variables remain constant. B suggests using Boyle's law then Charles's law separately, which is unnecessarily complicated and prone to error. The combined gas law handles all variables simultaneously, and the calculated value is incorrect. C references Gay-Lussac's law, which only relates pressure and temperature at constant volume—inappropriate here since volume clearly changes. This approach ignores the pressure change effects on volume. Study tip: For gas problems with multiple changing variables, always use the combined gas law first. Don't overcomplicate by applying individual gas laws sequentially unless specifically required. Remember that temperature must always be in Kelvin for gas law calculations.
A sealed, rigid container holds an ideal gas at a pressure of 1.5 × 10⁵ Pa and a temperature of 27 °C. The gas is then heated to a temperature of 127 °C. What is the new pressure of the gas?
Answer: 2.0 × 10⁵ Pa For an ideal gas in a sealed, rigid container, the volume (V) and the number of moles (n) are constant. According to the combined gas law (P₁V₁/T₁ = P₂V₂/T₂), this simplifies to Gay-Lussac's Law: P₁/T₁ = P₂/T₂. The temperatures must be in Kelvin. T₁ = 27 °C + 273 = 300 K. T₂ = 127 °C + 273 = 400 K. Rearranging for the final pressure, P₂ = P₁ × (T₂/T₁) = (1.5 × 10⁵ Pa) × (400 K / 300 K) = (1.5 × 10⁵ Pa) × (4/3) = 2.0 × 10⁵ Pa.
A flexible balloon is being inflated with more air at a constant temperature. The external atmospheric pressure is also constant. How is the volume $V$ of the balloon related to the number of air molecules $N$ inside it?
Answer: $V$ is directly proportional to $N$ . The ideal gas law is PV = Nk_B T. In this scenario, the pressure P is constant (equal to the external atmospheric pressure), the temperature T is constant, and $k_B$ is the Boltzmann constant. Rearranging the equation for V gives V = ( $k_B$ T / P) × N. Since the term in the parentheses is constant, the volume V is directly proportional to the number of molecules N. This is also known as Avogadro's Law.
Which change to a fixed mass of an ideal gas guarantees an increase in its internal energy?
Answer: Increasing its volume while keeping its pressure constant. The internal energy (U) of a fixed mass of an ideal gas is directly proportional to its absolute temperature (T). From the ideal gas law, PV=nRT, temperature is proportional to the product of pressure and volume (T ∝ PV). We are looking for a change that guarantees the product PV increases. A: The product PV could increase, decrease, or stay the same (e.g., P→2P, V→V/2 makes PV constant). B: If V increases at constant P, the product PV must increase. C: If P decreases at constant V, the product PV must decrease. D: The product PV could increase, decrease, or stay the same (e.g., P→P/2, V→2V makes PV constant). Therefore, only increasing the volume at constant pressure guarantees an increase in internal energy.
A weather balloon contains helium gas at ground level where the temperature is 288 K and pressure is 1.01 × 10⁵ Pa. As it rises to an altitude where the temperature is 216 K and pressure is 2.53 × 10⁴ Pa, the balloon's volume increases by a factor of 3.2. Assuming the balloon doesn't leak, what can be concluded about this process?
Answer: The process violates the ideal gas law, suggesting helium exhibits non-ideal behavior at altitude Using the combined gas law for constant amount of gas: (P₁V₁)/T₁ = (P₂V₂)/T₂. Rearranging: V₂/V₁ = (P₁/P₂)(T₂/T₁) = (1.01×10⁵/2.53×10⁴)(216/288) = (4.0)(0.75) = 3.0. The ideal gas law predicts a volume ratio of 3.0, but the observed ratio is 3.2. This 6.7% deviation suggests non-ideal behavior, possibly due to intermolecular forces becoming significant at lower pressure and temperature.
An ideal monatomic gas is held in a rigid container of fixed volume. The absolute temperature of the gas is quadrupled. By what factor does the root-mean-square (rms) speed of the gas molecules increase?
Answer: 2 The average kinetic energy of a gas molecule is directly proportional to the absolute temperature (T): $E_k$ = (1/2)m<c²> ∝ T. The root-mean-square speed (c_rms) is defined as $\sqrt{<c²>}$ . From the energy relation, <c²> ∝ T, which means c_rms ∝ $\sqrt{T}$ . If the absolute temperature T is quadrupled (T → 4T), the new rms speed will be proportional to $\sqrt{4T}$ = 2 $\sqrt{T}$ . Therefore, the rms speed increases by a factor of 2.
An ideal gas expands to four times its initial volume. The expansion happens in such a way that the product of pressure and volume remains constant ( $PV = \text{constant}$ ). What is the ratio of the final root-mean-square speed of the molecules to the initial root-mean-square speed?
Answer: 1 For a fixed amount of an ideal gas, the ideal gas law states PV = nRT. If the product PV remains constant during the process, then the temperature T must also remain constant. This describes an isothermal expansion. The root-mean-square speed of the molecules (c_rms) is directly related to the absolute temperature (c_rms ∝ $\sqrt{T}$ ). Since the temperature does not change, the rms speed also does not change. Therefore, the ratio of the final rms speed to the initial rms speed is 1.
A rigid container holds a mixture of two ideal gases: 2.0 mol of nitrogen at partial pressure 3.0 × 10⁴ Pa and 1.5 mol of oxygen at partial pressure 2.25 × 10⁴ Pa, both at 273 K. If the temperature is increased to 373 K while maintaining constant volume, what will be the new partial pressure of nitrogen?
Answer: 4.10 × 10⁴ Pa, since partial pressures increase proportionally with temperature For each component gas in a mixture, Gay-Lussac's law applies independently: P₁/T₁ = P₂/T₂. For nitrogen: P₂ = P₁(T₂/T₁) = (3.0 × 10⁴ Pa)(373 K/273 K) = (3.0 × 10⁴)(1.366) = 4.10 × 10⁴ Pa. Choice B incorrectly calculates using the wrong temperature ratio. Choice C incorrectly assumes pressure redistribution occurs between gases. Choice D incorrectly applies total pressure considerations when partial pressures are independent.
An ideal gas is confined to a cylinder by a piston. The gas is heated, causing it to expand from volume $V_1$ to $V_2$ in such a way that its pressure is directly proportional to its volume ( $P = kV$ for some constant $k$ ). If the initial temperature is $T_1$ , what is the final temperature $T_2$ ?
Answer: $T_1 (V_2/V_1)^2$ From the ideal gas law, T is proportional to PV for a fixed amount of gas (T ∝ PV). We are given that pressure is proportional to volume, P ∝ V. Substituting this into the temperature proportionality, we get T ∝ (V)V, which means T ∝ V². Therefore, the ratio of the temperatures is equal to the ratio of the volumes squared: T₂/T₁ = (V₂/V₁)². Solving for T₂ gives T₂ = T₁ (V₂/V₁)².
Container X holds an ideal gas of molar mass $M$ at pressure $P$ and temperature $T$ . Container Y holds a different ideal gas of molar mass (2M) at pressure $2P$ and temperature $4T$ . What is the ratio of the density of the gas in Y to the density of the gas in X ( $\rho_Y / \rho_X$ )?
Answer: 1.0 The ideal gas law is PV = nRT. Since the number of moles n = m/M (mass/molar mass) and density ρ = m/V, we can substitute these in. PV = (m/M)RT → P = (m/V)(RT/M) → P = ρ(RT/M). Rearranging for density gives ρ = PM/(RT). For container X, ρ_X = PM/(RT). For container Y, ρ_Y = (2P)(2M)/(R(4T)) = 4PM/(4RT) = PM/(RT). The ratio ρ_Y / ρ_X is (PM/(RT)) / (PM/(RT)) = 1.0.
A piston-cylinder assembly contains an ideal gas initially at 2.0 bar and 300 K. The gas undergoes two sequential processes: first, an adiabatic expansion where the volume triples, then an isothermal compression back to the original volume. If the adiabatic index γ = 1.4, what is the final temperature of the gas?
Answer: 162 K, calculated from the combined effect of both thermodynamic processes Process 1 (adiabatic): $TV^{γ-1}$ = constant, so T₁V₁^0.4 = T₂V₂^0.4. With V₂ = 3V₁: T₂ = T₁ $(V₁/V₂)^{0.4}$ = 300K $(1/3)^{0.4}$ = 300K(0.540) = 162K. Process 2 (isothermal): Temperature remains constant at 162K even though volume returns to V₁. The final state has the original volume but lower temperature and pressure. Choice A uses incorrect adiabatic calculation. Choice B incorrectly assumes return to initial state. Choice D uses wrong thermodynamic relations.
A fixed quantity of an ideal gas is cooled at constant pressure. What are the changes in the root-mean-square speed of the molecules and the density of the gas?
Answer: rms speed decreases, density increases The root-mean-square speed is proportional to the square root of the absolute temperature (c_rms ∝ $\sqrt{T}$ ). Since the gas is cooled, its temperature T decreases, so the rms speed must decrease. The process occurs at constant pressure for a fixed quantity of gas. According to the ideal gas law (PV=nRT), if T decreases while P and n are constant, the volume V must decrease. Density is defined as mass per unit volume (ρ = m/V). Since the mass of the gas is fixed and its volume decreases, its density must increase.

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