Qualification > Sciences

Physics CIE

(1/2) > >>

yasser37:
Hi
can anyone explain the table in november 10 paper 41
ques. 2
part 3

Sue T:
p to Q : work done on gas: its a decrease in volume at constant pressure so simply use :
delta W = P x delta V
           = 4 x 10^5 x (8-2) x 10^ -4
           = 240 J
so the increase in internal energy for that row would be -360 J
Q to R is obviously +720
now R to P you have 2 start with  the right hand side of the table :
now the total chng in internal energy for this cycle is zero ryt? since you end up where you started
so the entire right hand side column should add up 2 zero therefore:
-360 + 720 + x = 0
x = -360
now check the bottom row and make it add up
x + 480 = -360
x = -840
so r to P work done on gas is -840J

thats it

aldehyde1612:
Is that a)iii ... or b)iii

cause for a)iii ... Ideal gases dont have potential energy so internal energy is the sum of the kinetic energy of all the molecules, and kinetic energy is proportional to the temperature of the gas.? <Ek>= 3/2 kT .. right?

and b)iii. I dont know about that cause the marking scheme has the numbers all wrong.

Sue T:
he said table so i assumed the table...

aldehyde1612:
oh. lol. yeah you're right. sorry, I didnt read the post properly.
But thanks for that explanation.. I didnt know how to do that last row.  ;D

Navigation

[0] Message Index

[#] Next page