For the isobaric process as shown in the figure, find an expression for the first law that involves the pressure and volume. * P i f V

Needs Complete typed solution with 100 % accuracy.

can someone find delta S for an isothermal process joined with the middle (intermediate) staTe to the last (final) state. note the delta U does not equal 0

Prove that the “cycle” consisting of only two adiabatic steps (reversible expansion and then compression) will have no efficiency since no work can be produced. Will it violate the 2nd law of the thermodynamics? Clear explain your answer.

Please show work and use the correct unit. Thank you

Using the information provided (image), compare the change in entropy during the isothermal expansion from 1 L to 2 L of one mole of a rigid diatomic ideal gas and one mole of a rigid diatomic Van der Waals gas with b = 0.08 L/mol (ethanol). Are the changes in entropy the same for both or is one greater than the other?

= 2 is taken through a cycle as shown in the figure to the right, which is not necessarily drawn to scale. n moles of an ideal gas with adiabatic index Y The cycle proceeds as follows: 1– 2 is an isobaric compression to V2= Vo/2 2 – 3 is an isochoric process with P3 = 3P0 3 – 4 is an isothermal compression to V4 = Vo/4 4 – 5 is an isochoric heating process 5 – 1 is an adiabatic expansion back to Vo ЗРо Ро Vo/4 Vo/2 Vo (a) at constant pressure, What are Cv, the molar specific heat at constant volume, and Cp, the molar specific heat for this gas? Your answer should be in terms of n, R, T , or a subset of these quantities. (b) ( diagram) in terms of Po, and the temperatures T,, T3 , T4 , and T3 (the temperatures at points 2 – 5) in terms of To. All work must be shown. Find the unknown quantities P4 and P5 (which are the pressures at points 4 and 5 on the (c) Find the work done on the system in terms of n, R, T, , and pure numbers only for the 3, 3 → 4, 4 → 5, and 5 → 1. Make sure to show…