- #1
Correct for an ideal gas. This is just algebra ie:chuakoktong said:For the derivation of pv gamma using PV=nRT, i do know how to get it. However, i would like to try other concept to get it as well.
Sorry for my poor explanation.
At the moment, I would like to understand the below question to know whether i on the right track.
A)the internal energy of gas in a cylinder with volume V and pressure P can be written as
1/(γ-1)PV correct?
Not quite. The work done ON the gas is -PdV. The work done BY the gas is PdV.B)when compress the gas by a small volume dv, the work done on gas is P*dv?
Not quite. PdV is the work done BY the gas. The final internal energy is the sum of the work done ON the gas (-PdV) and the initial internal energy.C)The final internal energy is the sum of work done (P*dv)+initial internal energy(1/(γ-1)PV) correct?
You appear to be trying to avoid using calculus by approximating. To get an exact answer you have to use calculus.chuakoktong said:Dear Mason,
Consider a diatomic gas with pressure 1pascal undergo adiabatic compression by 0.1m^3
Using
P1V1^γ=P2V2^γ
(1)(1)^1.4=P2(0.9)^1.4
P2=1.1589pascal
This is what cause of my confusion
I been thinking may be is possible to get the same answer as above using the formula as mention below.
...
Compare value obtain from eq 1 and eq2, eq1 is closer to value of pressure obtain using PV^γ
approximation method always accompanied by error when comparing to the integral form.
However, using a formula does not match the original equation result in a bigger error?
At the moment i haven't try using smaller dv
You have to use the formula. I am not sure why you want do use an approximation. Just use PV^γ = constant.Also in adiabatic free expansion, most of it mention P1V1/T1=P2V2/T2
and T1=T2 lead to P1V1=P2V2
I am wondering also consider the case of there is no change in the internal energy of gas in adiabatic free expansion
5/2P1V1=5/2P2V2 which lead to P1V1=P2V2 also
I been thinking by using conservation of energy it should be able to hold a formula to predict any process but still the -Pdv sign and +pdv sign is confusing me
At the moment i will try to use a smaller dv to check the result for eq 1.
chuakoktong said:Dear Mason,
Thank for your fast reply.
The reason i advoid using calculus is because i trying to break down thing part by part for my further understanding.
I been thinking as far as things are interrelated, maybe same result/answer can also be obtained by using different way(i.e taking different path leading the same destination)
I am trying by means of calculus can the equation as below
5/2P1V1+Pdv=5/2P2V2
or
5/2P1V1-Pdv=5/2P2V2
integrated become P1V1^γ
An adiabatic process is a thermodynamic process in which there is no heat transfer between the system and its surroundings. This means that the system does not exchange thermal energy with its environment, and the process occurs without any change in temperature.
The adiabatic process is derived from the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In an adiabatic process, there is no heat added or subtracted, so the change in internal energy is equal to the work done by the system.
The formula for adiabatic process is PVγ = constant, where P is the pressure, V is the volume, and γ is the ratio of specific heats. This formula is also known as the adiabatic equation.
The main difference between adiabatic and isothermal processes is that in an adiabatic process, there is no heat transfer, whereas in an isothermal process, the temperature remains constant. In other words, an adiabatic process involves a change in temperature, while an isothermal process does not.
Some real-life examples of adiabatic process include the compression and expansion of gases in car engines, the compression of air in a bicycle pump, and the expansion of air in a hot air balloon. These processes occur without any heat transfer, making them adiabatic processes.