ive been stuck on this question for a while..so could someone explain how to sol

Question

ive been stuck on this question for a while..so could someone explain how to solve it?

The figure shows a reversible cycle through which 1.00 mole of a monatomic ideal gas is taken. Process bc is an adiabatic expansion, with Pb = 11.0 atm and Vb = 1.70 times 10-3 m3. For the cycle, find (a) the energy added to the gas as heat, (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle. Number Units J Number Units J Number Units J Number Units This answer has no units

Explanation / Answer

you are given the data

volume in L pressure in atm pV (in atm-L)

point a 1.70 ??? ???

point b 1.70 11.0 18.7

point c 13.6 ??? ???

Since b to c is adiabatic, you can use

pressure at b * volume at b ^ 5/3 = pressure at c * volume at c^5/3 or

11.0 * 1.70^5/3 = pressure at c * 13.6^5/3

so pressure at c = 0.34375 atm

Now

volume in L pressure in atm pV (in atm-L)

point a 1.70 0.34375 0.584375

point b 1.70 11.0 18.7

point c 13.6 0.34375 4.675

Now calculate delta U W and Q for each process using pV at each point:

a to b: delta U = (3/2) (18.7 - 0.584375) = 27.173

work = zero (constant volume)

heat = 27.173 atm-L entering gas

b to c: heat = zero (adiabatic)

delta U = (3/2) ( 4.675 - 18.7) = -21.04

work = 21.04 atm-L

c to a: delta U = (3/2) (0.584375 - 4.675) = -6.14

work = p deltaV = 0.34375 * (1.70 - 13.6) = -4.09

heat = -4.09 - 6.14 = -10.23 atm-L

(a) heat added to gas = 27.173 atm-L = 2753 Joules

(b) heat leaving gas = 10.23 atm-L = 1036 Joules

(c) net work = 21.04 - 4.09 = 16.95 atm-L = 1717 Joules

(d) eff = work done / heat added = 1717 / 2753 = 0.624

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