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Physics: Post your doubts here!

VelaneDeBeaute

VelaneDeBeaute

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Sandhya Mahat said:
Somebody please help me with this question. The answer is 20J but I don't know the process. Please answer along with the process.
50J of heat energy is supplied to a fixed mass of gas in a cylinder. The gas expands, doing 20J of work. Calculate the change in internal energy of the gas.
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I am not sure if the answer is 20J. I'm going to write down everything I think makes sense.
Change in Internal Energy = Change in Heat Energy + Change in Work done by the system on the gas
= 50J + (-20J) = 30J :confused:
 
Gémeaux

Gémeaux

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Sandhya Mahat said:
Somebody please help me with this question. The answer is 20J but I don't know the process. Please answer along with the process.
50J of heat energy is supplied to a fixed mass of gas in a cylinder. The gas expands, doing 20J of work. Calculate the change in internal energy of the gas.
Click to expand...
VelanedeBeaute did it correct i guess :/. Its like the 20J work is what *the gas does on the cylinder*, which is why it has to be subtracted.
If according to first law of thermodynamics, work done on gas is to be added, work done *by* the gas on the system is to be subtracted.
 
X

xyz!

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rheea.cool said:
can you tell me how to find absolute uncertainity anyone?
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hi sum1 had posted dis in sum other thread..i m pasting it here for u..c if it helps!
Example: Find lg R and its error and record in the table

R =231 +/- 10 (3s.f)
lg R should be to 3 sf

lg 231 = 7.85174904142
=> lg 231 = 7.852 (to 3 sf)

Notice in lg 7.852 is to 3 s.f and not 4. The numbers before the . do not count
E.g :
0.039 = 3 sf
3.430 = 3sf
3.4503 = 4sf
0.0002 = 4sf

Now the error in lg R = |lg Rmax - lg R|
Rmax = 231+10 = 241
Lg Rmax = Lg(241) = 7.91288933623 = 7.913 (to 3sf)

Therefore error in Lg R = |lg Rmax - lg R| = |7.913 - 7.852| = 0.061

So you record Lg R in the table as follows: 7.852 +/- 0.061

As you will see error above is to 3 s.f. Exceptionally in P5, the errors IN THE TABLE can be to more than 1s.f.

However when you will be calculating gradient and y-intercepts, you should bring the the error to 1s.f
Example for final answers of gradient and y-intercept:
gradient = 5.02 +/- 0.03 << Error to 1 s.f
y-intercept = 1.20 +/- 0.01 << Error to 1 s.f

So if it happens when u calculate gradient u get:
5.321 +/- 0.324 << This answer is not good and won't be accepted. You should work out the answer to the correct SF for the error. So you proceed as follows:

=>5.321 +/- 0.324 , error should be to 1 sf
=> 5.321 +/- 0.3 , now error in to 1 sf, but the gradient value is not good. Error is 1 SF but 1 D.P, so gradient should be to 1 D.P
=> 5.3 +/- 0.3 <<<<< That's the final answer for gradient


So to summarise:
> error in calculated values IN TABLE can exceptionally be to more than 1 sf.
> error in final answers for gradient or y-intercept should IMPERATIVELY be to 1 sf. Then you proceed as I explained above to get the final answer.


I suggest you go and work Nov 2009 P51. You will understand everything based on that paper. It's really the paper which has the most difficult data I've ever encountered.
 
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