The Ideal Gas Law
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AP Physics 2 › The Ideal Gas Law
A fixed amount of ideal gas is taken from state 1 to state 2 in a cylinder. The temperature is held constant at $300,\text{K}$ while the volume increases from $1.5,\text{L}$ to $3.0,\text{L}$. If $P_1=300,\text{kPa}$, then $P_2$ is
$450,\text{kPa}$ because $300\times(3.0/1.5)=450$.
$300,\text{kPa}$ because temperature is constant.
$600,\text{kPa}$ because pressure increases when volume increases.
$150,\text{kPa}$ because pressure is inversely proportional to volume.
Explanation
This question tests the ideal gas law. At constant temperature and amount of gas, the ideal gas law (PV = nRT) reduces to Boyle's Law: P₁V₁ = P₂V₂, showing that pressure and volume are inversely proportional. When volume doubles from 1.5 L to 3.0 L, pressure must be halved to maintain the constant product PV. Therefore, pressure decreases from 300 kPa to 150 kPa. Choice A incorrectly states that pressure increases when volume increases, revealing a fundamental misconception about the inverse relationship between P and V at constant temperature. When temperature is constant, remember that pressure and volume change in opposite directions by inverse factors.
An ideal gas in a cylinder with a movable piston is compressed slowly from $4.0,\text{L}$ to $2.0,\text{L}$ while the temperature and amount of gas are held constant. Initially the pressure is $100,\text{kPa}$. Compared to the initial pressure, the final pressure is
$50,\text{kPa}$ because pressure decreases when volume decreases.
$200,\text{kPa}$ because pressure is inversely proportional to volume.
$150,\text{kPa}$ because pressure increases by $50,\text{kPa}$.
$100,\text{kPa}$ because temperature is constant.
Explanation
This question tests the ideal gas law. The ideal gas law (PV = nRT) shows that when temperature and amount of gas are constant, pressure and volume are inversely proportional (P₁V₁ = P₂V₂). As volume decreases from 4.0 L to 2.0 L (halved), pressure must double to maintain the same product PV. Therefore, pressure increases from 100 kPa to 200 kPa. Choice A incorrectly states that pressure decreases when volume decreases, revealing a fundamental misconception about the inverse relationship between P and V. To solve ideal gas problems correctly, identify which variables are constant and apply the appropriate proportionality relationship.
A fixed amount of ideal gas is kept at constant temperature in a piston-cylinder device. The pressure is reduced from $300,\text{kPa}$ to $150,\text{kPa}$ while the amount of gas is constant. Compared to the initial volume, the final volume is
half as large because volume is proportional to pressure.
twice as large because volume is inversely proportional to pressure.
the same because temperature is constant.
four times as large because pressure was reduced by a factor of 2.
Explanation
This question tests the ideal gas law. At constant temperature with a fixed amount of gas, pressure and volume are inversely proportional according to Boyle's Law (P₁V₁ = P₂V₂). When pressure is halved from 300 kPa to 150 kPa, volume must double to maintain the constant product PV. Therefore, the final volume is twice the initial volume. Choice A incorrectly states that volume is proportional to pressure, confusing direct and inverse relationships. Remember that at constant temperature, P and V vary inversely—when one doubles, the other halves.
A rigid, sealed $2.0,\text{L}$ container holds an ideal gas at $1.0,\text{atm}$ and $300,\text{K}$. The amount of gas and volume are held constant while the gas is heated to $450,\text{K}$. Which statement correctly describes the final pressure?
It decreases to about $0.67,\text{atm}$ because temperature increased.
It stays at $1.0,\text{atm}$ because volume is constant.
It increases to about $1.5,\text{atm}$ because pressure is proportional to Kelvin temperature.
It increases to about $2.5,\text{atm}$ because $450-300=150$.
Explanation
This question tests the ideal gas law. The ideal gas law states that PV = nRT, where pressure (P), volume (V), amount of gas (n), and temperature (T) are related through the gas constant R. When volume and amount of gas are constant, pressure is directly proportional to absolute temperature (P₁/T₁ = P₂/T₂). Since temperature increases from 300 K to 450 K (a factor of 1.5), pressure must also increase by a factor of 1.5, from 1.0 atm to 1.5 atm. Choice D incorrectly adds the temperature difference (150 K) to the pressure, showing a misconception about how to apply proportional relationships. When solving ideal gas problems, always use absolute temperature in Kelvin and identify which variables remain constant to determine the appropriate relationship.
An ideal gas in a sealed, rigid container (constant volume and constant amount) is warmed from $27^\circ\text{C}$ to $127^\circ\text{C}$. Which statement correctly describes the final pressure compared with the initial pressure?
$P_2=\tfrac{400}{300}P_1$ because pressure is proportional to kelvin temperature
$P_2=P_1$ because rigid containers keep pressure constant
$P_2=\tfrac{127}{27}P_1$ because pressure is proportional to Celsius temperature
$P_2=\tfrac{300}{400}P_1$ because pressure decreases as temperature increases
Explanation
This question tests the ideal gas law. For constant volume and amount, pressure is proportional to absolute temperature: P₁/T₁ = P₂/T₂. First convert temperatures to Kelvin: T₁ = 27°C + 273 = 300 K and T₂ = 127°C + 273 = 400 K. Then P₂ = P₁ × (T₂/T₁) = P₁ × (400 K / 300 K) = (4/3)P₁. Choice A incorrectly uses Celsius temperatures directly (127/27), demonstrating the common misconception that temperature ratios work with any scale. Always convert to Kelvin before using temperature ratios in gas law calculations.
An ideal gas in a cylinder is kept at constant pressure by a movable piston. Initially $V_1=4.0,\text{L}$ at $T_1=250,\text{K}$. It is warmed to $T_2=300,\text{K}$ while pressure and amount of gas remain constant. Which statement correctly describes $V_2$?
It is $3.3,\text{L}$.
It is $54,\text{L}$ because $\Delta T=50^\circ\text{C}$.
It is $4.0,\text{L}$.
It is $4.8,\text{L}$.
Explanation
This question tests the ideal gas law. The ideal gas law PV = nRT shows how pressure, volume, temperature, and moles are interconnected. When pressure and moles are constant, volume is directly proportional to temperature: V₁/T₁ = V₂/T₂. With V₁ = 4.0 L, T₁ = 250 K, and T₂ = 300 K, we calculate V₂ = V₁T₂/T₁ = (4.0 L)(300 K)/(250 K) = 4.8 L. Choice D incorrectly uses the temperature difference in Celsius (50°C) rather than the ratio of absolute temperatures. Always convert to Kelvin and use ratios when applying Charles's Law: V₁/T₁ = V₂/T₂.
A flexible balloon contains ideal gas at constant temperature and external pressure. The balloon initially has $n_1=0.20,\text{mol}$ and volume $V_1=1.0,\text{L}$. Gas is added to reach $n_2=0.30,\text{mol}$ while temperature and pressure remain constant. Compared to $V_1$, what is $V_2$?
It is $0.67,\text{L}$.
It is $1.0,\text{L}$.
It is $1.5,\text{L}$.
It is $1.3,\text{L}$ because volume depends on Celsius temperature.
Explanation
This question tests the ideal gas law. According to PV = nRT, pressure, volume, temperature, and moles of gas are related in specific ways. When pressure and temperature are constant, volume is directly proportional to the number of moles: V₁/n₁ = V₂/n₂. With V₁ = 1.0 L, n₁ = 0.20 mol, and n₂ = 0.30 mol, we find V₂ = V₁n₂/n₁ = (1.0 L)(0.30 mol)/(0.20 mol) = 1.5 L. Choice D incorrectly suggests using Celsius temperature, which is irrelevant when temperature is constant. When P and T are constant, use Avogadro's Law: V₁/n₁ = V₂/n₂.
A rigid, sealed steel tank contains $1.0\ \text{mol}$ of an ideal gas at $300\ \text{K}$ and pressure $P_1$. The tank is heated to $450\ \text{K}$ while volume and amount of gas remain constant. Which statement correctly describes the new pressure $P_2$ compared to $P_1$?
$P_2=1.5P_1$
$P_2=\tfrac{450}{300+273}P_1$
$P_2=P_1+150\ \text{kPa}$
$P_2=\tfrac{2}{3}P_1$
Explanation
This problem tests understanding of the ideal gas law. The ideal gas law states that PV = nRT, where pressure (P), volume (V), and temperature (T) are related through the number of moles (n) and gas constant (R). When volume and amount of gas are constant, pressure is directly proportional to absolute temperature: P₁/T₁ = P₂/T₂. Since temperature increases from 300 K to 450 K, we get P₂ = P₁ × (450/300) = 1.5P₁. Choice D incorrectly uses Celsius temperature (450/(300+273)), showing the common misconception of not recognizing that temperatures are already in Kelvin. Always verify that temperatures are in absolute units (Kelvin) and identify which variables remain constant to determine the appropriate gas law relationship.
A sample of ideal gas is kept at constant pressure in a piston. Its temperature increases from $300,\text{K}$ to $360,\text{K}$, and no gas is added or removed. Which statement correctly describes how the volume changes?
The volume increases by $60%$ because $360-300=60$.
The volume increases by a factor of $360/300$ because volume is proportional to absolute temperature at constant pressure.
The volume stays the same because pressure is constant.
The volume decreases by a factor of $360/300$ because higher temperature compresses the gas.
Explanation
This question tests understanding of the ideal gas law. The ideal gas law shows that at constant pressure and amount of gas, volume is directly proportional to absolute temperature (Charles's Law): V₁/T₁ = V₂/T₂. The volume increases by the same factor as the temperature: V₂/V₁ = T₂/T₁ = 360 K / 300 K = 1.2, meaning volume increases by a factor of 360/300. Choice D incorrectly calculates the change as 60% by subtracting temperatures (360 - 300 = 60), showing a misconception about using temperature ratios rather than differences. Always use ratios of absolute temperatures, not temperature differences, when applying gas laws.
A fixed amount of ideal gas is in a cylinder with a frictionless piston. The gas is cooled from $T_1=500\ \text{K}$ to $T_2=250\ \text{K}$ while the external pressure is adjusted so the gas pressure remains constant. No gas enters or leaves. Which statement correctly describes $V_2$ compared with $V_1$?
$V_2=\tfrac{250-273}{500-273}V_1$ because Celsius temperatures must be used
$V_2=V_1$ because constant pressure implies constant volume
$V_2=\tfrac{1}{2}V_1$ because volume is proportional to kelvin temperature at constant pressure
$V_2=2V_1$ because volume increases when temperature decreases at constant pressure
Explanation
This question tests the ideal gas law. For an ideal gas with constant pressure and constant amount, PV = nRT simplifies to show that volume is directly proportional to absolute temperature: V₁/T₁ = V₂/T₂. Since temperature decreases from 500 K to 250 K (halved), the volume must also be halved: V₂ = V₁ × (250 K / 500 K) = ½V₁. Choice A incorrectly states that volume increases when temperature decreases, revealing a fundamental misconception about the direct relationship between V and T. When pressure and amount are constant, use V₁/T₁ = V₂/T₂ with Kelvin temperatures.