Consider the following two-step process. Heat is allowed to flow out of an ideal

Question

Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.5 atm. Then the gas expands at constant pressure, from a volume of 6.8 L to 10.0 L, where the temperature reaches its original value. See the figure. Calculate the total work done by the gas in the process. _________J Calculate the change in internal energy of the gas in the process. ___________ J Calculate the total heat flow into or out of the gas. _____________J out of the gas into the gas

Explanation / Answer

Given

the ideal gas at constant volume with pressure changing from 2.2 atm to 1.5 atm in the first stage later

the gas expands at constant pressure so the volume is chaning from v1 = 6.8 L to 10 L, reaches the original temperature so no change in temperature of the gas after reachng the point C

a) work done by the gas in the process is

during from a to b the volume change is zero so the work done w = P*DV =P*0 = 0 J

and from b to c is W2 = P*DV = 1.5*1.01325*10^5 (0.01-0.0068) J = 486.36 J

so the total work done W = 0+486.36 = 486.36 J

b)
we knwo that the change in internal energy is

   dU = ncV *dT
as there is no change in temperature of the gas so the change in internal enrgy of the gas =0

c) from the first law of thermodynamics

   dQ = dU+ dW

   dU=0

   dQ = dW = 486.36 J

so the heat is flowing into the gas

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