compressing an ideal gas

compressing an ideal gas

devesh9749

Member

Updated: Oct 26, 2024

Views: 10.0K

hello everyone,
since i am a crazy engineer so i had this crazy doubt that,

If we have an insulted container having a gas like say helium(which is close to ideal gas) and has a piston on top.
Now if we put a weight on the piston and allow the gas to be compressed to some volume will the gas get cooled?

can anyone please can tell whether my reasoning is correct or wrong?

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devesh9749

Member • Jul 1, 2008

someone please reply...

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Jeanius

Member • Jul 6, 2008

If you're looking at an ideal gas, then you're working with (phonetically) pivnert (PV=nRT).

Both n and R would remain constant throughout the push, so we really need to focus on the PV and T portions. What you'd really have to work out is the ratio between P and V in any specific case, as pressure will increase and volume will decrease. Think of a liquid oxygen tank expelling gas. The nozzle freezes as pressure decreases, right?

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Ernesha

Member • Apr 27, 2009

One of the postulates of Kinetic Molecular Theory is that ideal gases have no forces of attraction or repulsion between molecules. So when you compress an IDEAL pressure increases but with the same ratio Volume decreases so in the relation PV=nRT, T is not affected. But in the case of NON-IDEAL gases the force of attractions and repulsions ARE present and so when the gas is compressed it experiences heavy repulsions from the other molecules due to which its rms velocity increases and then K.E and ultimately its Temperature. But in case of IDEAL gases the rms velocity is not altered by the compression.

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Corvus

Member • Apr 27, 2009

The gas will heat up as you pressurize it

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g_rakesh2

Member • Apr 27, 2009

devesh9749
hello everyone,
since i am a crazy engineer so i had this crazy doubt that,

If we have an insulted container having a gas like say helium(which is close to ideal gas) and has a piston on top.
Now if we put a weight on the piston and allow the gas to be compressed to some volume will the gas get cooled?

can anyone please can tell whether my reasoning is correct or wrong?
hello devesh
PV=nRT is ideal gas law.

for ideal gas
By compressing it, you're changing the volume (decreasing) and the pressure (increasing).

It's possible that the combination of PV could remain constant, e.g. so that nRT remains a constant, so that T remains a constant.
have you ever seen air baloon seller.
baloon seller is having a tank in which a light gas is stored by copmressing it like helium if you touch that tank it will feel hot.
agreed ernesha!!!

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Rohan_sK

Member • Apr 28, 2009

The thermodynamic process in which the tempreature of the gas remains constant is called an ISOTHERMAL PROCESS. As all thermodynamic processes it is goverened by thr gas equation PV = mRT, where T is kept constant by providing a constant cooling reservoir around the system.
As for Devesh, where he wants to reduce the tempreature of the gas as well despite of compression, then the process will not be a standard process. It will need a calculated cooling procedure from outside, and the equation governing would be p1v1^n = p2v2^n, where value of 'n' would be the deciding factor.

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safwan

Member • Jul 26, 2009

Whether it's Constantly applied Pressure or variable mate...

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skipper

Member • Jul 26, 2009

As others have pointed out an ideal gas is isothermal (is an isotropic medium). An ideal gas can be compressed without heating, temperature stays constant and pressure increases linearly. Isothermal expansion and adiabatic compression are the keys to a successful heat engine, But all real engines lose heat.

A real gas does heat up because of nonlinearity. If an ideal gas existed it would be monatomic and smooth, with no dipoles of charge or mass, the spin would be irrelevant for each atom; in a real gas this isn't true except at some very general level that ignores quantum effects.
😎

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rajulucky2520

Member • Feb 16, 2011

hello davesh,
ur doubt can be clarified by viewing it in thermodynamics point of view.As u insulted the container no heat tranfer takes place then dQ=0,as gas is compressed work done is -Pdv. then apply fist law of thermodynamics for closed system....dQ-PdV=U2-U1..then it becomes -PdV=U2-U1 it implies U2<U1...internal energy is directly proportional to temperature....T2<T1...so the gas will gets cooled

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ISHAN TOPRE

Member • Feb 27, 2011

As PV=nrT is the law,if we vary V then P will change.Thereby maintaining T constant.So temperature will remain constant.
But as it is a piston cylinder arrangement with no room for extra volume,the gas will get heated up.

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PraveenKumar Purushothaman

Member • Mar 2, 2011

Hey basic Chemistry says, when molecules are tightly packed, they get heated up! So, obviously when you apply pressure, the gas heats up.
At the same time, when hydrogen or oxygen (non ideal gases) are compressed, on the other hand, they get super cooled! Basic Physics!

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Ramani Aswath

Member • Apr 6, 2011

devesh9749
If we have an insulted container having a gas like say helium allow the gas to be compressed to some volume will the gas get cooled?
Work has been done on the gas. Since the process as described is adiabatic, the gas will get heated.

Bioramani

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