# Determine initial and final temperatures, adiabatic expansion?

• snowicorn
In summary, the student attempted to solve for the initial temperature of an ideal diatomic gas expanding adiabatically using the equation PVγ= nRT. However, this equation is incorrect as it should be PV=nRT and during an adiabatic process, PVγ= constant. The student was also missing the value of γ for a diatomic gas in their calculation. The correct approach would be to first use the equation P1V1γ = P2V2γ to find the final pressure, and then use PV=nRT to solve for the initial temperature.
snowicorn

## Homework Statement

A 3.25 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1050 m^3 to 0.762 m^3. Initially the pressure was 1.00 atm

Determine the initial and final temperatures

PVγ= nRT

## The Attempt at a Solution

To get the initial temperature, I used the above equation

*note: converted atm to N/m^2

T = PVγ/nR = (1.013 *10^5 N/m^2)((0.1050 m^3)^1.4)/ (3.25 mol * 8.314 J/mol*K)

T = 159.8 K

Unfortunately that isn't the correct answer and the correct answer is actually 394 K. I'm pretty sure I'm going through the steps correctly. But no matter how I put the above into the calculator, I keep getting the wrong answer.

All help is appreciated! :)

snowicorn said:

## Homework Statement

A 3.25 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1050 m^3 to 0.762 m^3. Initially the pressure was 1.00 atm

Determine the initial and final temperatures

PVγ= nRT

## The Attempt at a Solution

To get the initial temperature, I used the above equation

*note: converted atm to N/m^2

T = PVγ/nR = (1.013 *10^5 N/m^2)((0.1050 m^3)^1.4)/ (3.25 mol * 8.314 J/mol*K)

T = 159.8 K

Unfortunately that isn't the correct answer and the correct answer is actually 394 K. I'm pretty sure I'm going through the steps correctly. But no matter how I put the above into the calculator, I keep getting the wrong answer.

All help is appreciated! :)

The equation marked with red is wrong.
PV=nRT for an ideal gas. And pVγ= const during an adiabatic process.

ehild

snowicorn said:

## Homework Statement

A 3.25 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1050 m^3 to 0.762 m^3. Initially the pressure was 1.00 atm

Determine the initial and final temperatures

PVγ= nRT

## The Attempt at a Solution

To get the initial temperature, I used the above equation

*note: converted atm to N/m^2

T = PVγ/nR = (1.013 *10^5 N/m^2)((0.1050 m^3)^1.4)/ (3.25 mol * 8.314 J/mol*K)

T = 159.8 K

Unfortunately that isn't the correct answer and the correct answer is actually 394 K. I'm pretty sure I'm going through the steps correctly. But no matter how I put the above into the calculator, I keep getting the wrong answer.

All help is appreciated! :)

You should first use P1V1γ = P2V2γ to evaluate for final pressure of the gas.

And , PVγ= nRT is wrong ! Its PVγ=constant. Also it will be correct to write PV=nRT...

Hint : What is γ for diatomic gas ?

## 1. What is meant by initial and final temperatures in adiabatic expansion?

In adiabatic expansion, the initial temperature refers to the temperature at the beginning of the expansion process. This is typically the temperature of the gas before it undergoes any change. The final temperature, on the other hand, is the temperature at the end of the expansion process, after the gas has expanded and reached equilibrium.

## 2. How is the initial and final temperature determined in adiabatic expansion?

The initial and final temperatures can be determined by using the ideal gas law, which states that the product of pressure and volume is directly proportional to the product of the number of moles of gas and the temperature. By manipulating this equation, the initial and final temperatures can be calculated.

## 3. What is the relationship between initial and final temperatures in adiabatic expansion?

In adiabatic expansion, the initial and final temperatures are inversely proportional to each other. This means that as the volume of the gas increases, the temperature decreases and vice versa. This relationship is known as Boyle's Law.

## 4. How does adiabatic expansion affect the temperature of a gas?

During adiabatic expansion, the gas expands without any exchange of heat with its surroundings. This causes the gas to do work and therefore, its internal energy decreases. As a result, the temperature of the gas also decreases.

## 5. What are some real-life examples of adiabatic expansion?

One common example of adiabatic expansion is the expansion of compressed air in a can when it is opened. The air inside the can undergoes adiabatic expansion, causing it to cool down and leading to the formation of condensation on the outside of the can. Other examples include the expansion of gases in a bicycle pump and the compression and expansion of air in a car engine.

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