Gas constants are critical in the study of gas laws and play a significant role in various scientific calculations. These constants allow chemists to relate the pressure, volume, and temperature of gases, facilitating the understanding of gas behavior in different conditions. Below are three diverse, practical examples that demonstrate the application of gas constants in real-world situations.
When discussing the behavior of gases, the Ideal Gas Law (PV = nRT) is one of the most fundamental equations. Imagine you have a balloon filled with helium gas at room temperature (25°C) and a pressure of 1 atmosphere. This scenario provides a clear context to apply the gas constant.
In this case:
Using the Ideal Gas Law:
\[ PV = nRT \]
\[ (1 \, atm) (2.5 \, L) = (0.1 \, mol)(0.0821 \, L·atm/(K·mol))(298 \, K) \]
\[ 2.5 = 0.1 imes 0.0821 imes 298 \]
\[ 2.5 = 2.466 \]
The calculated values are close, confirming that the gas behaves ideally under these conditions. This example illustrates how gas constants can help predict behavior in common scenarios.
Consider a scenario where a tire is inflated at sea level, where the atmospheric pressure is 1 atm. If you drive to a higher elevation where the pressure drops to 0.8 atm, understanding the gas constant’s role is crucial for predicting the volume change in the tire.
Using Boyle’s Law (P1V1 = P2V2), we can illustrate this change:
Applying Boyle’s Law:
\[ P1V1 = P2V2 \]
\[ (1 \, atm)(30 \, L) = (0.8 \, atm)(V2) \]
\[ 30 = 0.8V2 \]
\[ V2 = \frac{30}{0.8} = 37.5 \, L \]
As the pressure decreases, the volume of the gas inside the tire increases. This example highlights how gas constants are vital in practical applications, like tire inflation.
In a laboratory setting, chemists often need to determine the number of moles of a gas produced in a reaction. For example, consider a reaction that produces 5 L of carbon dioxide (CO2) at a pressure of 2 atm and a temperature of 350 K. This situation provides an opportunity to apply the Ideal Gas Law to find the number of moles.
Here, we use:
Rearranging the Ideal Gas Law to solve for n:
\[ n = \frac{PV}{RT} \]
\[ n = \frac{(2 \, atm)(5 \, L)}{(0.0821 \, L·atm/(K·mol))(350 \, K)} \]
\[ n = \frac{10}{28.735} \approx 0.348 \, mol \]
This calculation reveals that approximately 0.348 moles of CO2 were produced during the reaction. This example demonstrates the utility of gas constants in quantitative chemical analysis.