Problem 43 - Introductory Chemistry Atoms First

Question

Rewrite the ideal gas law solving for \(T\). Also show how all units cancel to leave you with just units of temperature.

Step-by-Step Solution

Answer

To rewrite the ideal gas law solving for temperature (T), we start with the equation \(PV = nRT\). By rearranging to isolate T, we get \(T = \frac{PV}{nR}\). Canceling out the units results in Kelvin (K) as the remaining unit, which confirms that the rewritten equation is correct.

Step 1: Write the ideal gas law equation

Write down the ideal gas law equation: PV = nRT

Step 2: Solve for temperature T

To solve for T, we need to isolate T on one side of the equation. Divide both sides by nR: \(T = \frac{PV}{nR}\) Now T is isolated and the equation is rewritten solving for temperature T.

Step 3: Cancel out units

To show how all units cancel to leave us with just units of temperature, let's break down the units for each variable: - P (pressure) has units of Pascals (Pa) or N/m² (Newtons per square meter) - V (volume) has units of cubic meters (m³) - n (number of moles) has units of moles (mol) - R (ideal gas constant) has units of J/(mol K) (Joules per mol per Kelvin) Now let's substitute these units into the equation we found in Step 2: \(\frac{N/m² * m³}{mol * J/(mol*K)}\) We can see that several units cancel out: - moles (mol) in both the numerator and denominator - meters (m) in both the numerator and denominator: m² in the denominator and m³ in the numerator, leaving m¹ (meters) in the numerator This leaves us with the following units: \(\frac{N * m}{J / K}\) We can further simplify the units by recognizing that the Newton-meter (Nm) is equivalent to Joules (J): \(\frac{J}{J / K} \) Finally, we see that the Joules (J) cancel out, leaving us with: K (Kelvin) This is the unit of temperature and confirms that our rewritten equation for T is correct: \(T = \frac{PV}{nR}\)

Key Concepts

Temperature Calculation

Temperature calculation using the Ideal Gas Law involves isolating the variable representing temperature. The Ideal Gas Law is expressed as:
\[ PV = nRT \]To focus on temperature, we need to solve for \( T \) by rearranging the equation. We do this by dividing both sides by the product of \( n \) and \( R \):

• Divide both sides by \( nR \): \[ T = \frac{PV}{nR} \]

This new form isolates \( T \) and directly links it to pressure \( P \), volume \( V \), and the product of moles and the gas constant. Understanding this step is crucial for calculating temperature using this law, as it confirms that temperature is proportionate to pressure and volume, and inversely proportionate to the product of moles and the gas constant.
This equation is applicable under the assumption of an ideal gas, where the gas particles are considered to have no interaction apart from elastic collisions. Therefore, being able to deduce \( T \) in terms of these variables helps us predict real-world behavior of gases under different conditions.

Gas Law Units

Understanding the units in the Ideal Gas Law is key. Let's break down the units involved in the equation \( PV = nRT \). The units come from each variable:

• Pressure \( P \): Usually measured in Pascals (Pa) or Newtons per square meter (N/m²)
• Volume \( V \): Expressed in cubic meters (m³)
• Amount of substance \( n \): Given in moles (mol)
• Ideal Gas Constant \( R \): Has units of Joules per mole per Kelvin (J/(mol·K))

When solving for temperature, these units must appropriately cancel out. Substituting these values into \[ T = \frac{PV}{nR} \] we get:\[ T = \frac{N/m² \times m³}{mol \times J/(mol\cdot K)} \].
Upon simplifying, the mole units cancel, leaving meters in the numerator and further simplification shows:\[ T = \frac{N \times m}{J/K} \].This defines the conversion of a Newton-meter to a Joule and illustrates that after cancellations, the unit for \( T \) is Kelvin.
Mastering these conversions and unit cancellations is crucial for resolving scientific problems correctly and confirming the validity of calculated results.

Kelvin Unit

The Kelvin unit ( K) is the SI unit for temperature and is pivotal in the Ideal Gas Law calculations. Unlike Celsius or Fahrenheit, Kelvin starts from absolute zero - the theoretical point where particles have minimal thermal motion.

• Kelvin provides a true scale and doesn’t have negative values because it starts from the fundamental thermal absence.
• One Kelvin increment is equivalent to one Celsius degree increment since they share the same scale difference, despite different starting points.

In gas law equations, using Kelvin ensures that temperature values are absolute, avoiding discrepancies that might occur with negative values in calculations involving zero or sub-zero temperatures in Celsius.
This is especially important in standard ideal gas calculations to ensure accuracy and consistency in results. Converting temperatures from Celsius to Kelvin involves adding 273.15 to the Celsius value, ensuring all temperature inputs are non-negative and universally comparable.