00:01
In the given question we are told that a carnot cycle is connected using an ideal diatomic gas.
00:06
Initially the gas at the temperature t equals to 25 degrees celsius.
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Pressure is given 100 kilo pascal.
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V1 is 0 .01 meter cube.
00:25
V2 0 .002 meter cube and pressure.
00:32
Is 263 .8 kilo pascal.
00:37
Here we have to find what is the maximum pressure of the case in the cycle above.
00:44
So here to find the maximum pressure of the gas for the process 1 to 3 1 to 2 that is isothermal.
00:58
We can write p2 v2 equals to p1 p1 so p2 will be 100 k 0 .01 divide by v2 that is 0 .002 so it will be 500 kilo pascal now for 2 to 3 that is antibiotic process p2 v2 raised to gamma equals to p3 v3 v3 raised to gamma so p3 will be 500 0 .002 divided by 0 .001 is to 1 minus 4.
01:41
So p3 will be 1319 .51 kiel pascal.
01:49
This is the maximum pressure of the gas in the above cycle.
01:58
Now to find the maximum temperature for process 2 to 3 .3 adiabatic process we can write t2 b raised to gamma minus 1 equals to t 3 b3 raised to gamma minus 1 so t 3 will be 298 0 .002 divide by 0 .01 raised to 0 minus 4 so t 3 that will be 393 .21 kelvin that is 120 .21.
02:39
1 so this is the maximum temperature in the cycle...