# Why doesn't pressure increase in an isothermal expansion of an ideal gas in a piston cylinder arrangement?

In a piston cylinder arrangement, the piston can be extended only if the pressure of the gas inside is higher than the atmospheric pressure.In case of isothermal expansion of ideal gas, initially the piston is at rest(gas pressure is equal to the atmospheric pressure) and as energy is given to the system the piston moves.Doesnt this mean the pressure of the gas increases above the outside pressure?

• Not if the volume increases. What does the ideal gas law predict? The gas is doing work in the surroundings, and heat is required to do this at constant temperature. May 23, 2018 at 13:50
• Does this mean, initially the pressure increases and pushes the piston and the increase in volume nullifies the increase in pressure(pressure is reduced back to its initial value) and the piston movement stops when the heat source is removed?(but in this case the pressure remains constant and doesn't reduce below the initial value as in case of isothermal expansion)
– user196272
May 23, 2018 at 14:10
• The movement stops if the temperature stops to increase. May 23, 2018 at 14:29
• But I think in case of an isothermal reaction the temperature remains constant. Mathematically I can understand that according to ideal gas law, if the temperature is constant then the increase in volume should be nullified by an equivalent decrease in pressure. But I couldn't understand how this process occurs physically.
– user196272
May 23, 2018 at 15:01
• At constant temperature, the pressure decreases as the volume increases. When heat is added to hold the temperature constant while the gas is doing work, the decrease in pressure is less than if the heat were not being added. At constant temperature, the amount of heat added precisely matches the work done by the gas. May 23, 2018 at 15:11

It is indeed quite correct that the pressure inside the cylinder increases above that of atmospheric conditions. However, the gradient of pressure (the pressure difference) is infinitesimally small. When we talk about isothermal conditions, we mean that the system is undergoing a thermodynamic process EXTREMELY SLOWLY, such that the system maintains a constant temperature relative to the surroundings. So, a pressure difference is inevitable, and that is precisely the factor that makes the piston move. If there were no pressure difference and if the process were isothermal, then two of the parameters governing a system, namely, the pressure and temperature, would be constant, thus making the third parameter—the volume—a constant as well.

The only reason we cannot perceive this infinitesimally small change in pressure is that no process is actually reversible and isothermal, as it would take an infinitely long period of time to undergo such process.

Also, I must add that for an ideal gas undergoing an isothermal process, Boyle's Law is applicable, which states that P×V = constant, and not just that P = constant. So, as the pressure increases infinitesimally, the piston moves outward, thus increasing the volume. Here, two counteracting processes are occurring one after another. This process continues as long as energy is supplied to the system.

• I think, If two counteracting processes occur one after another the pressure would be constant and the volume would increase(initially pressure would increase and then volume increases to nullify the effect i.e. pressure is brought back to its initial value.Then again pressure increases and this goes on).But in case of an isothermal expansion, pressure decreases with increase in volume.
– user196272
May 23, 2018 at 18:10
• Well, the counteracting processes do not occur in equal magnitudes. This is due to the physical property of the system. As the volume of the system increases, the piston moves outward. But due to the inertia of the system, the piston overshoots by a very small amount. AND THIS BECOMES THE NEW EQUILIBRIUM POSITION. Please see that the volume of the system has increased from it's initial state. Now, the same phenomenon continues, creating increase in volume and decrease in pressure. May 23, 2018 at 18:26
• But even if the counteracting processes occur at different magnitudes, the pressure will initially reduce and then remains constant at the reduced value(Say initially the gas is at the atmospheric pressure.Let B be a pressure value below the atmospheric pressure.When heated the pressure of the gas increases above the atmospheric pressure and pushes the piston.Due to inertia, piston moves a little more and the pressure is reduced to the value B below the atmospheric pressure.Now again due to heating, the pressure increases above atmospheric pressure, piston expands, pressure reduces to B .
– user196272
May 24, 2018 at 6:21
• In this case, the pressure reduces to B and remains constant(keeps returning to the value B) with increase in volume.
– user196272
May 24, 2018 at 6:53
• That is the issue here. It DOES NOT REMAIN AT B anymore. The new equilibrium state becomes (P1,V1,TO), where V1 is little more than what is expected from Ideal Gas Equation. This leads to decrease in pressure P1 below the expected value. If the energy supply were to be cut off at this point, then the atmospheric pressure would pressurize the piston down to the state, as expected from Ideal Gas Equation. But since energy is being provided continuously, the infinitesimal increase in pressure and consequent more increase in volume gradually proceeds on and on. May 24, 2018 at 8:34