A 1.00 -mol sample of an ideal gas (γ=1.40) is carried through the Carnot cycle described in Fig

A 1.00 -mol sample of an ideal gas (γ=1.40) is carried...

Key Concepts

Thermodynamic Efficiency

Thermodynamic efficiency, particularly in the context of a Carnot engine, quantifies the effectiveness of converting heat into work. It is defined as the ratio of work output to heat input and, for a Carnot cycle, is solely determined by the reservoir temperatures. This concept underscores how the temperature limits of the reservoirs dictate the theoretical maximum efficiency of an engine.

Isothermal Process

An isothermal process is one that occurs at a constant temperature. In this type of process, any work done by or on the system is exactly balanced by heat exchange with the surroundings. Isothermal processes are essential in the Carnot cycle for maintaining a constant temperature during heat intake and rejection, allowing for the direct computation of work and heat transfer along these segments.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings. In thermodynamic cycles, such as the Carnot cycle, the adiabatic steps are characterized by changes in temperature and volume without heat transfer, and they follow the relation PV^? = constant (where ? is the heat capacity ratio). This process is key to determining the intermediate states in a cycle.

Ideal Gas Law

The ideal gas law is a fundamental relation in thermodynamics expressed as PV = nRT, which links pressure (P), volume (V), temperature (T), and the amount of gas (n) through the gas constant (R). This law is instrumental in analyzing the state variables of a gas throughout different processes within a cycle and is especially important when dealing with thermodynamic cycles involving gases.

Carnot Cycle

The Carnot cycle represents an idealized thermodynamic cycle that is reversible and consists of two isothermal processes (heat transfer at constant temperature) and two adiabatic processes (no heat transfer). This concept is crucial because it provides the maximum possible efficiency that any heat engine can achieve operating between two temperatures, serving as a standard for evaluating real engines.

Transcript

00:01 Hello everyone in this problem there is one mole of ideal gas number of mole is one of ideal gas having gamma to be 1 .4 is carried out through the card notode cycle as soon in the figure at point a pressure is 25 atmosphere that is 25 into 1 .0 113 10 to the power 5 pascal and temperature at a is given 600 kelvin.

01:03 Pressure at c point is given 1 atmosphere.

01:11 So it would be 1 .013 into 10 to the power 5 pascal and temperature at c point is 400 kelvin.

01:31 In the part a we have to find pressure in volume at different.

01:40 Points a, b, c and d.

01:49 In b part we have to calculate net work done in the cycle, net work done in one cycle.

02:04 And in c part we have to calculate the efficiency of the cycle.

02:12 Let us start solving.

02:13 It is a, it having lot of calculation, very lengthy calculation.

02:19 Please be patient calculation for a using the ideal gas equation pv is called to nrt volume we can calculate nrt by p since we know the pressure and temperature at point a and c so volume at a you may calculate number of mole is 1 our ideal gas constant is 8 .314 and temperature at a is given 600 kelvin and pressure is 25 atmospheric but we are using in pascal so from here volume at a point you will get 1 .97 into 10 to the power of minus 3 meter cube so this is the one of the answer now volume at c 1 mole are 8 .314 temperature at c is given 400 atmospheric pressure is one atmospheric so on solving it volume at c you will get 32 .8 into 10 to the power of minus 3 meter cube so we have obtained volume at a volume at c and pressure already we know since a to b is isothermal process and v2 c is adiabatic process so for isothermal process av, b can write p -a -v -a -2 -b and for adiabetic process b to c you can write combining these two equations 1 in 2 b can write when we substitute all these values we will get the volume at b point let us here since it is ratio so directly we can put in atmospheric pressure volume at c you know to the power 1 .4 divided by 32 .8 into 10 to the power minus 3 whole to the power 1 divided by gamma minus 1 that is 1 .4 minus 1 so on solving it volume at b point we will get 11 .9 into 10 to the power of minus 3 meter now similarly we can calculate the pressure at b point pressure and volume at b point you can form the equation same way b to c adiabetic process and c2 d is is so for b2c adiabetic that is pv v v v to the b power gamma isccccccccccc to the b vc to the power gamma and for c to d isothermal process we can write pcvc is equal to combining 3 and 4 we can write volume at d will be pa upon pc into va to the power gamma minus 1 sorry gamma divided by whole to the power whole to the power 1 upon gamma minus 1 on substituting the value volume at d point we will get 25 upon 1 into 1 .97 into 10 to the power minus 3 hold to the power 1 .4 1 .97 into 10 .97 into 10 to the power 1 up on gama minus 1 so on solving it volume at the point you will get 544 into 10 to the power minus 3 meter cube now since av is isothermal process so for a v process you can write p a v a to b so from here pressure at b point you will get substitute the value 25 atmosphere volume at a 32 .8 into 10 to the power minus 3 meter cube divided by 5 .44 into 10 to the power minus 3.

11:44 So pressure at b point you will get 4 .14 atmosphere.

11:57 Now for cd isothermal process we can write pressure at c into volume at c to be called to pressure at d into volume at d.

12:22 So pressure at d point we will get now substitute the volume.

12:32 Pressure at c is 1 atmosphere.

12:36 Volume at c is 32 .8 into 10 to the power minus 3 meter cube and that at d point it is 5 .44 10 to the power minus 3.

12:53 So on solving pressure at d you will get 6 .03 atmospheric pressure...

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