Adiabatic process of monatomic gas problem

[athula kumara]

Homework Statement

By how much does the temperature of an ideal monatomic gas change in an adiabatic process in which 4.0kJ of work are done on each mole of gas?

By how much does the temperature of an ideal diatomic gas (with molecular rotation but no vibration) change in an adiabatic process in which 4.0kJof work are done on each mole of gas?

Homework Equations

The Attempt at a Solution

 

[athula kumara]

hi my problem is this information enough to do it. I don't know volume or pressure

 

[ehild]

hi my problem is this information enough to do it. I don't know volume or pressure

How is the internal energy of an ideal monoatomic gas related to the temperature?
How much does the internal energy of one mole ideal gas change in an adiabatic process if 4kJ work is done on the gas?

 

[athula kumara]

U=3/2 NkT, ok I got it.
I'm going to substitute values in this equation
thanks

 

[HalfLostAndSearching]

In an adiabatic process, there is no exchange of heat between the system and its surroundings. This means that the change in temperature, denoted as ∆T, is related to the work done, W, by the following equation:

∆T = - W / nCv

Where n is the number of moles of gas and Cv is the specific heat at constant volume. For a monatomic gas, Cv is equal to 3/2R, where R is the gas constant. Therefore, for the monatomic gas, the change in temperature can be calculated as:

∆T = - (4.0kJ/mol) / (1 mol) * (3/2R) = -6.0K

For a diatomic gas, the specific heat at constant volume, Cv, is equal to 5/2R. Therefore, the change in temperature for a diatomic gas can be calculated as:

∆T = - (4.0kJ/mol) / (1 mol) * (5/2R) = -8.0K

It is important to note that in an adiabatic process, the change in temperature is dependent on the specific heat at constant volume, which varies for different types of gases. Therefore, the change in temperature for a diatomic gas will be greater than that of a monatomic gas, as seen in the calculations above.