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Boundless Physics
Temperature and Kinetic Theory
Ideal Gas Law
Physics Textbooks Boundless Physics Temperature and Kinetic Theory Ideal Gas Law
Physics Textbooks Boundless Physics Temperature and Kinetic Theory
Physics Textbooks Boundless Physics
Physics Textbooks
Physics
Concept Version 10
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Problem Solving

With the ideal gas law we can figure pressure, volume or temperature, and the number of moles of gases under ideal thermodynamic conditions.

Learning Objective

  • Identify steps used to solve the ideal gas equation.


Key Points

    • Write down all the information that you know about the gas and convert the known values to SI units if necessary.
    • Choose a relevant gas law equation that will allow you to calculate the unknown variable, and substitute the known values into the equation. Then calculate the unknown variable.
    • The general gas equation only applies if the molar quantity of the gas is fixed.

Terms

  • SI units

    International System of Units (abbreviated SI from French: Le Système international d'unités). It is the modern form of the metric system.

  • ideal gas

    A hypothetical gas whose molecules exhibit no interaction and undergo elastic collision with each other and with the walls of the container.


Full Text

The Ideal Gas Law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behavior of many gases under many conditions, although it has several limitations. It is most accurate for monatomic gases at high temperatures and low pressures.

The ideal gas law has the form:

$PV= nRT$,

where R is the universal gas constant, and with it we can find values of the pressure P, volume V, temperature T, or number of moles n under a certain ideal thermodynamic condition. Typically, you are given enough parameters to calculate the unknown. Variations of the ideal gas equation may help solving the problem easily. Here are some general tips.

The ideal gas law can also come in the form:

$PV= NkT$,

where N is the number of particles in the gas and k is the Boltzmann constant.

To solve the ideal gas equation:

  1. Write down all the information that you know about the gas.
  2. If necessary, convert the known values to SI units.
  3. Choose a relevant gas law equation that will allow you to calculate the unknown variable.
  4. Substitute the known values into the equation. Calculate the unknown variable.

Remember that the general gas equation only applies if the molar quantity of the gas is fixed. For example, if a gas is mixed with another gas, you may have to apply the equation separately for individual gases.

Example

Let's imagine that at the beginning of a journey a truck tire has a volume of 30,000 cm3 and an internal pressure of 170 kPa. The temperature of the tire is 16∘C. By the end of the trip, the volume of the tire has increased to 32,000 cm3 and the temperature of the air inside the tire is 40∘C. What is the tire pressure at the end of the journey?

Tire Pressure

Tire pressure may change significantly during the operation of the vehicle. This is mostly due to the temperature change of the air in tires.

Solution:

Step 1. Write down all the information that you know about the gas: P1 = 170 kPa and P2 is unknown. V1 = 30,000 cm3 and V2 = 32,000 cm3. T1 = 16∘C and T2 = 40∘C.

Step 2. Convert the known values to SI units if necessary: Here, temperature must be converted into Kelvin. Therefore, T1 = 16 + 273 = 289 K, T2 = 40 + 273 = 313 K

Step 3. Choose a relevant gas law equation that will allow you to calculate the unknown variable: We can use the general gas equation to solve this problem: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.

Therefore, $P_2 = \frac{P_1 \times V_1 \times T_2}{T_1 \times V_2}$.

Step 4. Substitute the known values into the equation. Calculate the unknown variable:

$P_2 = \frac{170\times 30,000 \times 313}{289 \times 32,000} = 173~kPa$.

The pressure of the tire at the end of the journey is 173 kPa.

Note that in Step 2 we did not bother to convert the volume values to m3. In Step 4, pressure appears both in the numerator and denominator. In this case the conversion was not necessary.

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