Physics: Post your doubts here!

VelaneDeBeaute

VelaneDeBeaute

Messages
1,882
Reaction score
1,331
Points
173
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...
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

Messages
1,558
Reaction score
2,428
Points
273
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!

Messages
99
Reaction score
27
Points
28
rheea.cool said:
can you tell me how to find absolute uncertainity anyone?
Click to expand...
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.
 
You must log in or register to reply here.
Top