A gas undergoes a cyclic process, starting at point A with a volume of 0.02m³ going to B (isochoric heating),then to C (isothermal expansion) and finally back to A. If the heat added during isothermal expansion is 1000J and the heat rejected during isochoric heating is 500J, Calculate the net work done by the system. 12 class physics Numericals First law of thermodynamics numericals Physics class 12 numericals New Sindh Text Physics numericals
00:00Numerical number 5 of unit 16 first law of thermodynamics is a gaze undergoes a cyclic
00:16process starting at point A with a volume of 0.02 m3 going to be isochoric heating then
00:30to see isothermal expansion and finally back to A. If the heat added during isothermal expansion
00:42is 1000 Joule and the heat rejected during isochoric heating is 500 Joule. Calculate the net work
00:54done by the system. So first of all we will make data, sab se pahele hum data banayenge
01:09is heat added means q1 is given as 1000 Joule and heat rejected q2 is given as
01:39500 Joule. So we have to find work done. For solution or calculation, in case of isothermal
02:04expansion heat supplied is equal to work done by the system. So we have to find work done
02:18is change in heat energy delta q is equal to q1 minus q2. So work done will be equal to q1 minus q2. So work done will be equal to as q1 is 1000 Joule minus q2 is
02:25q2 is 500 Joule. So work done is 500 Joule. So work done will be obtained as 1000 Joule minus q2 is 500 Joule. So work done will be obtained as
02:32work done will be equal to Q1 minus Q2. So, work done will be equal to as Q1 is 1000 Joule
02:45minus Q2 is 500 Joule. So, work done will be obtained as 1000 minus 500, it will be
02:58500 Joules. So, this is our final answer means work done will be 500 Joule.
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