Q.34

Q34. 1 mole of a diatomic gas is contained in a piston. It gains 50.0J of heat and work is done on the surrounding by system is - 100 J. Thus, 

( A )   t h e   g a s   w i l l   c o o l   b y   2 . 41 °   ( B )   t h e   g a s   w i l l   h e a t   b y   2 . 41 °     ( C )   t h e   g a s   w i l l   c o o l   b y   3 . 61 °     ( D )   t h e   g a s   w i l l   h e a t   b y   3 . 61 °  

The solution is as follows:


T = CQ = 75R-100+50  = -2.41

The correct option is (a) the gas will cool by 2.410.

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