The Ideal Gas Law and Combined Gas Law: Understanding and Applications
The Ideal Gas Law: A Universal Equation for Gases
Ideal gas law
The ideal gas law is the equation of state of a hypothetical ideal gas which relates the pressure, volume, temperature, and amount of substance in a gas.
The ideal gas law is a mathematical relationship that connects four key variables describing a gas: pressure ($P$), volume ($V$), temperature ($T$), and the number of moles ($n$):
$$PV = nRT$$
Here:
- $P$: Pressure (measured in pascals, Pa),
- $V$: Volume (measured in cubic meters, m³),
- $n$: Number of moles of gas,
- $R$: Universal gas constant ($8.31 \, \mathrm{J \, mol^{-1} \, K^{-1}}$),
- $T$: Temperature (measured in kelvin, K).
Key Insights from the Ideal Gas Law:
- Pressure and Volume Relationship:
- Compressing a gas (increasing $P$) reduces its volume ($V$), while reducing the pressure allows the gas to expand, assuming constant temperature and number of moles.
- Temperature and Volume Relationship:
- Heating a gas increases its volume because the particles move faster and exert more outward force.
- Amount of Gas:
- Adding more gas molecules (increasing $n$) increases the pressure or volume, depending on the situation.
- Always convert temperature to kelvin by adding $273.15$ to the Celsius value before using the ideal gas law.
- Kelvin is the absolute temperature scale required for gas law calculations.
A 0.500 m³ tank contains $2.00 \, \mathrm{mol}$ of oxygen gas at a temperature of $300 \, \mathrm{K}$. What is the pressure inside the tank?
Solution
- Write the ideal gas law: $PV = nRT$
- Rearrange for pressure:
$$P = \frac{nRT}{V}$$ - Substitute the known values:
$$P = \frac{(2.00)(8.31)(300)}{0.500}$$ - Calculate:
$$P = 9,972 \, \mathrm{Pa} \, \text{or approximately} \, 10.0 \, \mathrm{kPa}$$
- In this example, we calculated the pressure inside a tank using the ideal gas law.
- Notice how each unit (moles, kelvin, and cubic meters) aligns with the units of $R$.
- This consistency is critical for accurate results.
The Combined Gas Law: Relating Initial and Final States of a Gas
While the ideal gas law is useful for a single set of conditions, many situations involve a gas changing state.
A balloon might expand as it rises to higher altitudes where the pressure decreases.
The combined gas law relates the initial and final states of a gas:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ where:
- $P_1, V_1, T_1$: Initial pressure, volume, and temperature,
- $P_2, V_2, T_2$: Final pressure, volume, and temperature.
Key Assumptions:
- The amount of gas ($n$) remains constant.
- Temperatures must always be in kelvin.
- Use the combined gas law when two or more variables (pressure, volume, temperature) change simultaneously.
- It’s especially useful for predicting gas behavior in dynamic conditions, such as weather changes or altitude shifts.
A weather balloon has a volume of $32.0 \, \mathrm{dm^3}$ at sea level ($P_1 = 100.0 \, \mathrm{kPa}$, $T_1 = 298 \, \mathrm{K}$). At an altitude of $4500 \, \mathrm{m}$, the pressure drops to $57.7 \, \mathrm{kPa}$, and the temperature is $273 \, \mathrm{K}$. What is the balloon’s new volume?
Solution
- Write the combined gas law:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ - Rearrange for $V_2$:
$$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$$ - Substitute the known values:
$$V_2 = \frac{(100.0)(32.0)(273)}{(57.7)(298)}$$ - Calculate:
$$V_2 \approx 51.2 \, \mathrm{dm^3}$$
Using the combined gas law, we predicted the balloon’s volume at a higher altitude. The balloon expands because the pressure drops more significantly than the temperature.
Applications: Determining Molar Mass from Experimental Data
- The ideal gas law can also be used to calculate the molar mass of a gas.
- Molar mass ($M$) is defined as the mass of one mole of a substance.
- If you know the mass ($m$) of a gas, its volume ($V$), temperature ($T$), and pressure ($P$), you can determine its molar mass using the following formula: $$M = \frac{mRT}{PV}$$
A $2.00 \, \mathrm{dm^3}$ sample of an unknown gas at STP weighs $3.88 \, \mathrm{g}$. Determine its molar mass.
Solution
- Recall that at STP ($P = 100.0 \, \mathrm{kPa}$, $T = 273.15 \, \mathrm{K}$), $R = 8.31 \, \mathrm{J \, mol^{-1} \, K^{-1}}$
- Write the formula:
$$M = \frac{mRT}{PV}$$ - Substitute the known values:
$$M = \frac{(4.40)(8.31)(273.15)}{(100.0\times 10^3)(2.00 \times 10^{-3})}$$ - Calculate:
$$M \approx 44.0 \, \text{g mol}^{-1}$$
- At STP, the molar volume of any ideal gas is $22.7 \, \mathrm{dm^3 mol^{-1}}$.
- This can simplify calculations involving volume and moles.
Units and Conversions: A Critical Step
- Volume:
- Use cubic meters ($m^3$) in calculations.
- Convert from:
- $1 \, \mathrm{dm^3} = 10^{-3} \, \mathrm{m^3}$,
- $1 \, \mathrm{cm^3} = 10^{-6} \, \mathrm{m^3}$.
- Pressure:
- Use pascals ($1 \ \mathrm{Pa} = 1 \, \mathrm{N \ m}^2$).
- Common conversions:
- $1 \, \mathrm{kPa} = 10^3 \, \mathrm{Pa}$,
- $1 \, \mathrm{atm} = 101.3 \, \mathrm{kPa}$.
- Many students forget to convert temperature to kelvin or use incorrect units for pressure and volume.
- Always double-check your units before solving gas law problems.
- Can you calculate the volume of a gas at a new temperature and pressure using the combined gas law?
- What steps would you follow to determine the molar mass of a gas?
