An ideal diatomic gas, m a cylinder with a movable piston, undergoes the rectang

Question

An ideal diatomic gas, m a cylinder with a movable piston, undergoes the rectangular cyclic process shown in Figure 1.10(b). Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are "frozen out." Also assume that the only type of work done on the gas is quasistatic compression-expansion work. For each of the four steps A through D, compute the work done on the gas, the heat added to the gas, and the change in the energy content of the gas. Express all answers in terms of P1, P2 , V1, and V2 . (Hint: Compute AU before Q, using the ideal gas law and the equipartition theorem.) Describe in words what is physically being done during each of the four steps; for example, during step A , heat is added to the gas (from an external flame or something) while the piston is held fixed. Compute the net work done on the gas, the net heat added to the gas, and the net change in the energy of the gas during the entire cycle. Are the results as you expected? Explain briefly.

Explanation / Answer

I will explain in the case B as it has more number of stepsyou follow the same lines and try to analyse case(A) in case ofdifficulty please let me know Line A represents constant volume process, piston remains inthe same position Hence external work done on the system. dw = 0J And from the graph pressure is increasing according to lineA for that kinetic energy of gas molecules should increase onlypossible when we supply heat to the system. But from first law of thermodynamics dQ = dU + dW dQ = dU therefore all the heat supplied will increase the internalenergy of the system Line B represents change of volume from V1 tovolume V2 at constant pressure process here volume increases that means work is done by the systemand is positive. for this we need to supply energy to the system some of part of this energy will be used to do external workto increase its volume against external pressure and some part to increase its internal energy that is its kineticenergy. Here nothing will become zero dQ = dU + dW = dU + PdV Line C represents constant volume process, piston remains inthe same position Hence external work done on the system. dw = 0J And from the graph pressure is decreasing according toline c for that kinetic energy of gas molecules should decreaseonly possible when  heat rejects by the system. But from first law of thermodynamics dQ = dU + dW dQ = -dU therefore all the heat rejected is equal to decreasein the internal energy of the system Line B represents change of volume from V2 tovolume V1 at constant pressure process here volume decreases that means work is done on thesystem and is negative. Hence heat will be rejected by the system some of part of this energy will from external work some part from its internal energy that is its kineticenergy. Here nothing will become zero dQ = dU + dW = dU + PdV all together it is a cyclic process net change in internalenergy, dU = 0 but heat supplied is equal to the net work done by thesystem. here work done is equal to the area of PV graph dW =(P2-P1)(V2-V1) hence if we apply first law of thermodynamics to thiscyclic process we get dQ = dW =(P2-P1)(V2-V1) Hence external work done on the system. dw = 0J And from the graph pressure is decreasing according toline c for that kinetic energy of gas molecules should decreaseonly possible when  heat rejects by the system. But from first law of thermodynamics dQ = dU + dW dQ = -dU therefore all the heat rejected is equal to decreasein the internal energy of the system Line B represents change of volume from V2 tovolume V1 at constant pressure process here volume decreases that means work is done on thesystem and is negative. Hence heat will be rejected by the system some of part of this energy will from external work some part from its internal energy that is its kineticenergy. Here nothing will become zero dQ = dU + dW = dU + PdV all together it is a cyclic process net change in internalenergy, dU = 0 but heat supplied is equal to the net work done by thesystem. here work done is equal to the area of PV graph dW =(P2-P1)(V2-V1) hence if we apply first law of thermodynamics to thiscyclic process we get dQ = dW =(P2-P1)(V2-V1) Line B represents change of volume from V2 tovolume V1 at constant pressure process here volume decreases that means work is done on thesystem and is negative. Hence heat will be rejected by the system some of part of this energy will from external work some part from its internal energy that is its kineticenergy. Here nothing will become zero dQ = dU + dW = dU + PdV all together it is a cyclic process net change in internalenergy, dU = 0 but heat supplied is equal to the net work done by thesystem. here work done is equal to the area of PV graph dW =(P2-P1)(V2-V1) hence if we apply first law of thermodynamics to thiscyclic process we get dQ = dW =(P2-P1)(V2-V1)

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