00:01
Hi everyone in this problem as the conducting valves is fixed off the container is fixed worked on by the case on the left part during the process is zero for left part let pressure bp and volume vv number of moles and it is given one mode and initial temperature is t1 so we can write pb by 2 is called to n r t1 so p b by 2 is called to r t1 so t1 is equal to p b upon 2 r 4 right hand side number of moles initial temperature t 2 so using the gas equation we can write p b by 2 is called to nr, t2.
02:44
So t2 you will get pv upon 4r.
02:49
So you will get t1 and t2, c part.
02:57
Let final temperature be t.
03:10
So total number of moles, 1 mole plus 2 mole, that is 3 mole.
03:23
Using gas equation, pv is called to nrt.
03:29
So temperature to be pv upon nr, that is pv upon 3m ,000.
03:44
For rhs using first law of thermodynamics mix, delta q is equal to delta u plus delta w but delta w here is 0.
03:58
So delta q is called to delta u since delta w is zero...