1 mole of an ideal gas (at 25°C and 1 atm pressure) is placed in a cylinder with a piston. Ambient temperature is 25°C and atmospheric pressure is 1 atm. Now if heat is applied to the system (ideal gas) in an isothermal process, its temperature will not rise but pressure and volume will change. How is this possible? Why does it keep the temperature constant instead of pressure? Is this isothermal process actually possible?
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3 Answers
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The gas pushes the piston outwards, doing work against the piston. In an isothermal process for an ideal gas the work is done by the gas at the same rate that heat flows into the gas, so, according to the first law of thermodynamics: Gain in internal energy = Heat flow into gas – Work done by gas, there will be no change in the internal energy, $U$, of the gas and (since the gas is ideal and $U=nc_vT$) no change in its temperature.
On a molecular scale, the heat flow into the gas occurs because gas molecules hitting the cylinder wall bounce back with, on average, more kinetic energy than on approach, owing to the enhanced vibration of the molecules of the slightly higher temperature wall. But gas molecules hitting the piston that is moving outwards bounce back with less kinetic energy than on approach.
"Is this isothermal process actually possible?" For the process to be reversible, the temperature difference between the cylinder walls and the gas must approach zero, and the speed of motion of the piston must approach zero, so reversible isothermal expansion is indeed an ideal, and is not strictly practicable.
answered Oct 30, 2022 at 14:48
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No, as far as I know isothermal process are not for heat engines that have high rpm. Isothermal process is slow or quasi-static. By doing so, it can keep the temperature of initial point of process, otherwise temperature fall or rise during expansion or compression, and it do that by with contact in heat reservoir to exchange heat.
It keeps temperature constant instead of pressure so heat added can be converted to work done, otherwise it is used in adiabatic if internal energy used that heat. There is no change in internal energy because temperature is constant during process.
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If your system is very small w.r.t. the dimension of the room where you're doing your experiment, we can think at the room as a heat reservoir external environment that is slightly affected by the heat exchanged with the system and can be kept at constant temperature.
You can think at a isothermal process as a very slow process, to give enough time to the system to be always in thermal equilibrium with the external environment, the big room, that is kept approximately at constant temperature.
