![]() But how can that be, if one instant earlier, the pressure inside the cylinder was much higher? What roughly happens is that, immediately adjacent to the piston, a small pressure dilitation region begins to form within the cylinder in which the pressure is equal to the external pressure. Now, if the piston is massless, that must mean that the pressures on both sides of the piston (after release) must be equal to one another (by Newton's 2nd law). We assume that the external pressure on the piston is held constant throughout the expansion. At time zero, the massless piston is released, and the volume of gas within the cylinder begins to increase. ![]() Problem statement: initially the gas pressure within the cylinder is higher than the pressure outside the cylinder, and the two regions are initially separated by a massless piston which is held in place. #Workdone with differenet pressure how to#Here is how to dope out what is happening: The basic answer is that, during the expansion, the pressure within the cylinder is not uniform spatially, and is lower at the piston than away from the piston. But, by looking at the problem in greater detail, you can get a better understanding of what is happening. This is a question which has driven thermo students crazy over the generations, usually without a satisfactory explanation. This is, of course not generally true of other state variables. Of the state variables we can usually claim that the volume change in the system = volume change in the surroundings. ![]() So the system pressure changes during the work exchange.įor the system the work, w = ∫pdv if the system pressure can be defined at all times during the process.įor the surroundings w = ∫pdv again, but since the pressure is constant w = p∫dv = p(v 2-v 1) With regard to you specific question it is assumed that the surroundings are so large that there is no change to the pressure during the work exchange. Sometimes one calculation is easier and sometimes one calculation is impossible. This gives us two opportunities to calculate the value of exchange variables - either form the system or from the surroundings - because they refer to the same quantity. It is a prime assumption of thermodynamics that if a quantity defined by one of these exchange variables is passed across the boundary it is the same on both sides of the boundary.įor example, work done by surroundings = work received by system (and vice versa) They do not describe the state of anything. These variables account for exchanges across the boundary. This would be pretty boring and useless if there was no interaction between the system and its surroundings so further variables are defined by the system process. Thermodynamics divides the universe into two parts.įor both system and surroundings state variables refer exclusively to either the system or the surroundings.Īs you are probably aware, state variable changes are only dependent upon the start and end points, not the intermediate path. Here is a viewpoint that may help with this and other questions. Hello bipolarity, I see you are asking lots of questions, that's good. I have heard some people say that the internal pressure of the gas over the process is undefined since the process is not quasi-static, but that should make the work done impossible to be calculated rather than just use external pressure which has no reason to be used in the problem? After all, if you calculate the work done on something, you should use the force or pressure applied on that object, or not the other way around, correct? So why do we use external pressure instead of internal pressure when calculating the work done by the gas? ![]() This is what I don't understand is that when you calculate the work done by the gas, you use the external pressure and not the gas pressure. Suppose initially that P>P and the gas expands until P=P In order for the gas to expand, P must be greater than P. ![]() Suppose in a piston in which there is a gas, the gas exerts pressure P on surroundings, whereas the surrounding exerts a pressure P on the gas. ![]()
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