One mole helium is enclosed in a cylinder with a moveable piston By placing the

Question

One mole helium is enclosed in a cylinder with a moveable piston By placing the cylinder in contact with various reservoirs and also insulating it at the proper times, the helium preforms the cycle shown in the figure below Process AB is isothermal. BC is adiabatic. CD is isobaric. and DA is isochoric Compute the internal energy change, heat transferred, and work performed for each portion of the cycle and total amount of each of these quantities for entire cycle Assume helium to be an ideal gas and C_p=20.8 J/mole K and C_v=12.5 J/mol.K.

Explanation / Answer

This system returns to its original state after completing a series of changes, then it is known that a cycle is completed.In a cyclic process the initial and the final state is same. As the internal energy U of the system depends only on the state of the system so in a cyclic process the net change of internal energy U will be equal to zero i.e. U = 0.
Further From the first law U = 0 = q + w, So, q=-W for complete cyclic system.
1) Process AB- isothermal: For the isothermal expansion of ideal gas U = 0.
so, q=-W= RTln(V2/V1) = 0.0831 x 19.2 x ln (0.005/0.002) = 1.46 J
2) Process BC-Adiabatic: Adiabatic is an isoleted system so no heat flow takes place. So, q=0 and U=-W = Cv (Tb – Ta). Here Tb & Ta are equal (19.2K) so U=-W = 0.
3)Process CD- isobaric: In this process no heat flow takes place. So, U = 0. and -q=W = R T ln (V4 / V3)= 1.46J…
4) Process DA- Isochoric process: q=0 So,U=W= CvT = Cv (T2 – T1). Again Here Tb & Ta are equal (19.2K) so U=-W = 0.

So total Internal energy Energy is = 0.0
and heat and work are in opposite direction for this cyclic process so q= W= 0 J.

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