A flask with 2.50 L of gas is under 3.00 atm of pressure. the gas is heated and the volume increases to 5.00 L. what is work done by the gas? (in J)

Question

aravindg · Answer

is temp constant??

anonymous · Answer

i think so , it says nothing about temp

aravindg · Answer

wel i suppose then it is isothermal expansion

aravindg · Answer

use the formula for work done in osthermal expansion to get the result

anonymous · Answer

is it, deltaH=deltaE + deltanRT

anonymous · Answer

or deltaE = q+w?

aravindg · Answer

oh not that

aravindg · Answer

use the formula :below http://en.wikipedia.org/wiki/Isothermal_process

aravindg · Answer

under calculation of work done

anonymous · Answer

w=pdv?

aravindg · Answer

nooo!!

aravindg · Answer

look under calculation of work done

anonymous · Answer

i dont have mol

aravindg · Answer

oh srry  then i suppose u jst need to do this:
work=PdeltaV

aravindg · Answer

u get that?

anonymous · Answer

can u solve it and show me

aravindg · Answer

srry i am a bit busy

jfraser · Answer

work done at constant temp is $$w = -P \Delta V$$

anonymous · Answer

how do i use this equation

anonymous · Answer

Just plug the numbers in. $$w=-3.00atm(5.00L-2.50L)$$

anonymous · Answer

the answer is suppose to be 759.8J ,  but i think ur right and the worksheet may be wrong

anonymous · Answer

well right now it's in units of liters times atms. We convert that using the equation$${101.31Joules \over L* atm} \times 7.5  L* atm = 759.8J$$ the 101.31 number you should have memorized.

anonymous · Answer

Always assume a work sheet is right. 99% of the time they are.

anonymous · Answer

lol thank you so much!

aravindg · Answer

hmm..... isnt that what i just said :P

anonymous · Answer

SO SIMPLE QUESTION

anonymous · Answer

WORK  OF EXPANSION=pRESSURE*CHANGE IN VOLUME NOW ACCORDING TO UR QUESTION 
W=P DELTA V 
W=3(5-2.50)
W=3*2.5
W=7.5