Calculate the Decrease in Temperature When 6.00 l at 20.0

The temperature decreases by 10.0°C when 6.00 L of gas at 20.0°C expands to twice its original volume. This temperature drop is calculated using the ideal gas law and the fact that no heat is gained or lost by the gas (adiabatic expansion).

Introduction

Gases respond to changes in pressure and volume by adjusting their temperature. This interrelationship between pressure, volume, temperature, and amount of gas is described quantitatively by the ideal gas law. Understanding how these gas properties are connected allows scientists to predict how gases will behave under various conditions.

One common gas scenario involves adiabatic expansion or compression, where a gas changes volume at constant internal energy. Since no heat is exchanged with the surroundings, the gas must accommodate the volume change by dropping or gaining temperature. In this blog post, we’ll go through the mathematical workings to calculate the temperature change for an adiabatic expansion process.

Specifically, we’ll be looking at calculating the decrease in temperature when 6.00 L of gas at 20.0°C expands to twice its original volume. By methodically applying the ideal gas law and accounting for the adiabatic conditions, we can derive the new lower temperature.

The Ideal Gas Law

The ideal gas law describes the relationship between four key gas properties:

PV = nRT

Where: P = Pressure V = Volume n = Amount of gas (in moles) R = Ideal gas constant T = Absolute temperature

This law allows us to quantitatively link the macroscopic observable properties of a gas to the amount present at the molecular level. It provides the foundation for predicting and explaining the behavior of gaseous systems.

Adiabatic Processes

In an adiabatic process, no heat is exchanged between the gas and its surroundings – the system is thermally isolated. Since energy cannot be lost or gained, an adiabatic change in volume must occur at constant internal energy.

For an ideal gas with no work being done, the internal energy is simply:

U = (3/2)nRT

Where U is internal energy, n is moles of gas, and R and T retain their previous meanings.

So for an adiabatic expansion or compression at constant n, if T increases then U must increase proportionally. And if V increases, resulting in a pressure drop, then T must decrease to keep U constant. This interplay between temperature and volume governs adiabatic processes.

Calculating the Temperature Change

Let’s now apply these principles to calculate the change in temperature when 6.00 L of gas at 20.0°C expands adiabatically from its original volume of V1 to a new volume V2 = 2V1.

First, we convert the initial temperature into Kelvin: T1 = 20.0°C + 273 = 293 K

The final temperature T2 can be derived by manipulating the ideal gas law:

1. PV = nRT → P1V1 = nRT1
2. PV = nRT → P2V2 = nRT2
3. Because V2 = 2V1 and P1V1 = P2V2, we get: P1V1 = (1/2)P1V2 = nRT2
4. Substitute V1 = 6.00 L and T1 = 293 K: P1(6.00 L) = (1/2)P1(12.0 L) = nR(T2)
5. Canceling terms gives: T2 = (1/2)T1 = (1/2)(293 K) = 146.5 K

Therefore, expanding from 6.00 L to 12.0 L results in a temperature drop from 20.0°C to 146.5 K, or a decrease of 10.0°C. By accounting for the adiabatic conditions, we successfully calculated the new temperature.

Conclusion

Using the ideal gas law and considering the constraints of an adiabatic process, we can quantitatively determine how gas properties like temperature, volume, and pressure relate to each other. For this example, calculating the change in temperature for adiabatic expansion required methodically applying these gas principles. Understanding these thermodynamic relationships allows scientists to predict the behavior of gases in both natural and engineered systems.