An ideal gas (1.0 mol) is the working substance in an engine that operates on the cycle shown i

An ideal gas (1.0 mol) is the working substance in an...

Key Concepts

Adiabatic Process

An adiabatic process is a thermodynamic process in which no heat is exchanged between a system and its surroundings. For reversible adiabatic processes, the change in the system’s internal energy results solely from work done by or on the system. This concept is key in analyzing cycles where rapid compression or expansion occurs, influencing the temperature and pressure characteristics of the working substance.

Reversible Process

A reversible process is an idealized process that occurs infinitely slowly, allowing the system to remain in thermodynamic equilibrium at all stages. These processes are theoretical constructs that enable maximum work extraction from a system or minimum work required for a process. They serve as benchmarks in thermodynamics, particularly useful in deriving maximum efficiency limits and understanding ideal cycle behavior.

Ideal Gas

An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. It is characterized by the ideal gas law, PV = nRT, which relates its pressure, volume, temperature, and amount of substance. This concept is fundamental in thermodynamics because it allows for simplified calculations and modeling of gas behavior under various conditions.

Degrees of Freedom and Molecular Structure

The degrees of freedom of a molecule, which depend on its structure (monatomic, diatomic, or polyatomic), determine how energy is distributed among translational, rotational, and vibrational motions. This concept is important when evaluating thermodynamic properties such as heat capacities and internal energy, as these properties influence the behavior of the substance during different thermodynamic processes.

Engine Efficiency

Engine efficiency in thermodynamics is defined as the ratio of the work output to the heat input. It serves as a measure of how effectively an engine converts heat energy into work. In cyclic processes, calculating efficiency involves analyzing the net work done over a cycle and accounting for heat transfers during various processes, which is essential for comparing real engines to idealized models.

Transcript

00:01 Now here in this question we are given that an idle gas that has number of moles equals to 1 is used as a working substance in an engine that operates on this cycle that you can see on this pv diagram and the process b to c this process is reversible adiabetic process and the process d to a is also reversible adiabetic process so we can write that these two processes are isentropic process because isentropic process means reversible adiabetic so isentropic process now in the a part we have to tell whether the idle gas use is monoatomic diatomic or polo atomic so for that if you consider the process d to a this process d to a and this process is isentropic process or you could say the reversible adiabetic process so for adiabetic process we can write that pressure at point d times the volume at point d to the power gamma is equal to pressure at point a times volume at point a to the power gamma now pressure at point d this is pressure at point d so pressure at point d we have p .0 divided by 32 so this is p .0 divide by 32 multiplied with volume at point d so volume at point d we have have 8 times v0 so this will be 8 times v0 to the power gamma is equals to now pressure at point a we have p .0 so that will be p .0 times the volume at point a we have v0 so v0 to the power gamma so from here this p .0 and p.

01:56 Not will cancel it out this one and this one and v0 and v0 will also go to cancel it out and we'll get that 8 to the power gamma divide by 32 is equal to and here it would become one only so that will be equals to 8 to the power gamma is equals to 32 and this 32 can be written as 2 to the power sorry 8 to the power 5 by 3 we have written 32 as 8 to the power 5 by 3 we can write this way also now if you compare that the bases are same so power must be same so gamma comes out to be as 5 by 3 and we know that gamma is 5 by 3 for mono atomic gases so therefore the gas used here it is mono atomic so this is the answer for the part now next we're going to solve the b part of the problem now in the b part we have to calculate the efficiency of the engine efficiency we know it's basically it is the ratio of work output divide by the heat input so work output can be written as qh that is heat supplied minus ql heat rejected divide by heat input we have qh so from here efficiency comes out to be as 1 minus qh sorry this is ql divided by qh now to calculate efficiency we have to calculate ql that is the heat rejected and then qh that is the heat now he should know that heat is added during the process a to b so this is qh here heat is added this is qh and heat is rejected during the process c to d so this is ql now qh is equal is equal as equals to n times the process is a constant pressure process so it will be n times cpv here to take multiplied with delta d now for mono or gas cp we know it is 5 by 2 r and delta t we have temperature at point b minus temperature at point a now if you use the idle gas equation we can write the temperature at point a will be equals to pressure at point a times the volume at point a divide by nr similarly the temperature at point b can be written as pressure at point b times the volume at point b divide by nr.

04:59 Now if you take the ratio of tb over t a, so t b over t a it will be equals to, so pressure at point b times the volume at point b divide by pressure at point a times the volume at point a.

05:18 Now it will be equals to pressure at point b we have p0 so that will be p now volume at point b is two times v0 so two times v0 divide by pressure at point a so pressure at point a we have p .0 so this will be p not times the volume at point a so volume at point a we have v0 so this will be v0 so the ratio comes out to v as equals to 2 now from this equation this is qh equals to n times 5 by 2 r i can take t a common so temperature at point a we have taken common and we get t b over t a minus 1 and again by using the idle gas equation we can write that p ava is equals to n r t a so this is n r t a can be written as p a so let us plug it here 5 by 2 remains as it is and this can be written as pressure at point a times the volume at point a multiplied with tb divided by t a minus one now we can plug it the values of five by two now pressure at point a this point a which is it is p not so this will be p not and volume at point a is v0 so v0 multiplied with t b over t a we have calculated as two so two minus one is one so we'll get this value therefore we have obtained that qh is equal to 5 by 2 p .0 v0.

07:08 Now next we're going to find the expression for ql that is the amount of heat rejected during the cycle.

07:15 So heat rejected that is ql it will be equals to...

Please give Ace some feedback