Question

show your work please!

Next two problems have to do with the following situation. A sample of monatomic ideal gas is changed from state A to PIsotherms state D. The gas first undergoes an isobaric process A - E and then an isothermal process E - D. Note that relevant parameters (pressures and volumes) are directly given on the graph. (Note that these problems are 2 atm -- -- directly taken from your HW assigned during a class. We have also solved an almost identical problem in class. See your class notes and my lecture slides.) 1 atma 4L SL16L 17. (2 points) Find the total work done by the gas. 18. (3 points) Find overall change in intemal energy of the gas.

Answer 1

For mono atomic gas

7 = 5/3

PA= PE = 2 atm = 2 * 101325 Pa, Pp=1 atm

VA = 41 = 4*10-3 m3. VE = 8 L = 8*10-3m. Vp = 16 L = 16*10-3m

TE = TD

The process undergoes from state A to state D.

17.

Work done by gas in process  AE is

WAE = PAVE-VA) = 2 * 101325 * (8 – 4) * 10-3 = 810.6 J

Work done by gas in process ED is
Vp WED = nRTE In = PEVE In = (2 * 101325*8 * 10-3) In 16 * 10-3 (8*10-3 = 1123.7J
Total work done by gas
W = 810..6 + 1123.7 = 1934.3J

18.

Change in internal energy of gas in process AE,

R AUAE = nCydT = n PAVE - VA) -TA) = 1 7-1

AUAE == 2 * 101325(8 – 4)10-3 -= 1215.9.) 5/3 - 1

Change in internal energy of gas in process ED ,

AUED = nCv (TD -TE) = 0

Overall change in internal energy of gas is

AU = 1215.9 +0= 1215.9 J