work transfer question

Feb 2020
4
0
United Kingdom
A gas in a vertical piston-cylinder has a volume 0.5m^3 and a temperature of 400k. The piston has a mass of 50kg and a cross-sectional area of 0.2m^2. As the gas cools down to atmospheric temperature the volume decreases to 0.375m^3. Neglect friction and assume atmospheric pressure is 100kPa. Determine the work transfer during the process.

The answer is -12.81kJ but I cannot arrive at this solution.

I tried using the relationship that pV = k (some constant) then subbing p = k/V into the integral of p with limits v1 and v2, then attempt to solve this integral, but I do not get the right answer.

In general, I am struggling with textbook questions relating to work and energy in my introductory course to thermodynamics already... if someone could point me in the right direction to assist my understanding in this module, that would be greatly appreciated....
 
Apr 2015
1,219
349
Somerset, England
I am suprised if that was the exact wording of question.

I suggest you start off by sorting out in your own mind the correct sequence of events, inlcuding what does work on what because the statement
the work transfer is -12.81 kJ is meaningless.

Work is done by something on something else, it is not transferred.

On the other hand, heat is transferred.

To start you off, I am guessing that what the question is saying is that

As the gas cools heat is transferred from the gas to its surroundings and also that since the gas volume is reduced, work is done on the gas.

So the change in internal energy of the gas will be the agebraic sum of the loss of heat and the work done on the gas, giving due regard to the sign.

You do not know the atmouspheric temperature, but do know its pressure.

You can calculate the distance moved by the piston from the data given.
Is the piston horizontal or vertical ? so is work done against the atmouspheric pressure or gravity or both?
 
Feb 2020
4
0
United Kingdom
I am suprised if that was the exact wording of question.

I suggest you start off by sorting out in your own mind the correct sequence of events, inlcuding what does work on what because the statement
the work transfer is -12.81 kJ is meaningless.

Work is done by something on something else, it is not transferred.

On the other hand, heat is transferred.

To start you off, I am guessing that what the question is saying is that

As the gas cools heat is transferred from the gas to its surroundings and also that since the gas volume is reduced, work is done on the gas.

So the change in internal energy of the gas will be the agebraic sum of the loss of heat and the work done on the gas, giving due regard to the sign.

You do not know the atmouspheric temperature, but do know its pressure.

You can calculate the distance moved by the piston from the data given.
Is the piston horizontal or vertical ? so is work done against the atmouspheric pressure or gravity or both?
the piston is shown to be vertical. so work is done with both gravity and atmospheric pressure (or in other words, against the gas).

and i use the volume to find the change in length of the piston?
 
Apr 2015
1,219
349
Somerset, England
and i use the volume to find the change in length of the piston?
Yes can you do this?
 
Apr 2015
1,219
349
Somerset, England
So is the situation as in my sketch ?

(and have you now calculated d ?)piston1.jpg
 
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Feb 2020
4
0
United Kingdom
Yes can you do this?
Yes, i have calculated the answer to 12,806.6kJ

I have found that L1 and L2 are 2.5m and 1.875m respectively. I can also change dV to AdL: \(\displaystyle \int^{V_1}_{V_2} p\;dV \) can be written as \(\displaystyle \int^{L_1}_{L_2} p\;AdL \). I then just subbed p = F/A, put the data in and solved the integral.

ty for your help.
 
Apr 2015
1,219
349
Somerset, England
OK so it is worth comparing my alternative method.

In my diagram the piston moves down a distance given by 0.2d = (0.5 - .375) = 0.625 m as you have.

The weight of the piston is constant as is the downward pressure of the atmousphere.

So the total downward force = weight + pressure x area = (50 x g)/1000 + 100 x 0.2 kN = (0.49 + 20) kN

This force moves its point of application 0.625 m so the work done = 20.49 x 0.625 = 12.81 kJ
 
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