The figure shows a reversible cycle through which 100 mole...

Transcript

00:01 Now here you see this is a pv diagram and the cycle is a reversible cycle through which a mono -atomic gas is taken and the number of mole in we have one and the volume at point c is the volume at point c vc vc is equals to eight times va so it is eight times the volume at point a and the volume at point a and b is same that means this process is an iso -coly process or you could say a constant volume process and process c to a is constant pressure process and the process b to c is a diabetic process now there are four parts in the equation in the a part where we have to calculate the energy added to the gas as heat so let me show it on the diagram in the process a to b heat is added this is qh now in the b part we have to calculate the energy leaving the gas as heat so in this process the energy is leaving so this is ql now in the b in the c part we have to calculate the net work done by the gas and finally the efficiency of the cycle so we're going to start with the a part where we have to calculate the value of qh that is the amount of heat added to the gas as heat now heat is added during the process a to b so qs will be equals to ncv delta t here delta t is a change in temperature between points a to b so it will be n cv final temperature we have t b minus initial temperature we have t a now we know that for mono atomic gas cv is equals to 3 over 2 r so we can plug it cv as 3 over 2 r where r is universal gas constant so it will be equals to three divided by two n remains as it is and r t b minus t a now here we don't know the value of t b and t a we can't find the heat supplied so for that we're going to apply the idle gas equation so let me write here ideal gas equation now at point a we know that the pressure at point a times the volume at point a is equals to n times r times temperature at point a so from here temperature at point a will be equals to p a va va divide by nr and pa that means the pressure at point a so pressure at point a is pa and volume at point a is va so we know that va equals to vb so we can write vb here it's one and the same thing similarly we can find that temperature at point b will be equals to pb vb vb divide by n r now we have obtained the expression for t a and t b we can plug it here so q will be equals to this is q supplied energy supplied so three divide by two nr now tb we have this much so p b vb divide by nr minus t a we have this much so p a v a divide by nr now here we see that this nr will entered out this one and finally we get the expression for qs to be 3 divided by 2 and v a actually here v a it was vb because va and vb are same so we can write vb here and we can take vb common so vb is common times pressure at point b minus pressure at point a so this expression that we have obtained but pressure at point a we are not given in the question and pressure at point b we are given to be as 10 atmospheric pressure it is given in the equation that pressure at point b is 10 atm so we can plug it here volume at point b also know we know that because the volume at point a we are given in the equation to be sorry at point a this is here so this volume we are given to be as 10 to the power minus 3 meter cube and volume at point c is eight times the volume at point a so that will be eight times 10 to the power minus 3 mera cube now let us calculate the pressure at point a first so pressure at point a pressure at point a will be equals to now here we see that now here we see that the process c to a is isobaric process that means the pressure remains constant so we can say that pressure at point b will be to pressure at point c is equal to now if you consider the process b to c so let me write a pressure process b to c and process b to c is a diabetic process and for adiabetic process we can write p c divide by p b is equals to v b divide by vc to the power gamma so from here the pressure at point c will be equals to p b times this vector so that will be p b times v b divided by vc to the power gamma and for monotomic gas gamma we know it is 5 by 3 now we can put the values so pressure at point b we have 10 atmospheric so 10 atmospheric means times 10 1 .01 3 times 10 to the power 5 and the ratio of vb over vc will be 1 by 8 because we are given that vc is 8 times of vb and vb means we can also write 8 times vb like this.

06:57 This is va so it can also be written as 8 times vb because va and vb are same.

07:02 So the ratio of vb over vc will be 1 by 8 to the power 5 over 3.

07:08 So from here we'll get the pressure at point a to be equals to 3 .167 times 10 to the power 4 pascal now we have obtained all the values we can plug it here to calculate the heat supplied so qs will be equals to 3 divided by 2 times vb now vb it is given in the equation to be 10 to the power minus 3 meter cube multiplied with pb minus pa so pb we have 10 atmospheric pressure that means 10 times 1 .013 times 10 to the power 5 pascal minus pa we have calculated as 3 .167 times 10 to the power 4 pascal and from here we can we can calculate by using a calculator and we'll get the value of qs to be 1 .47 times 10 to the power 3 jule so this is the heat supplied during the cycle and the answer for the a part.

08:26 Now next we're going to solve the b part of the problem where we have to calculate the heat rejected or you could say this value ql...

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