Molar Heat capacity

Classical Mechanics Level 5

1 1 Mole of an Ideal monoatomic gas undergoes a process depicted in the figure. At some point A of the P − V P-V diagram line joining the A to origin makes an angle 30 30 degrees with the V V axis and the tangent at point A makes an angle 120 120 degrees with the V V axis.

Find the instantaneous molar heat capacity of gas at point A in c a l m o l − 1 K − 1 cal mol^{-1} K^{-1} Take R = 2 c a l m o l − 1 K − 1 R=2 cal mol^{-1} K^{-1}


The answer is 2.

Ishan Tarunesh by Ishan Tarunesh , 23, India

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1 solution

Rohit Shah
Mar 27, 2015

Easy problem ! F r o m g r a p h , P = V 3 . . . ( 1 ) a n d d P d V = − 3 . . ( 2 ) S i n c e P V = n R T P d V + V d P = n R d T N o w u s i n g 1 a n d 2 a n d s o m e s i m p l e m a i n p u l a t i o n , P d V = − n R d T 2 d U = 3 2 n R d T d q = d U + d W d q = n R d T M o l a r h e a t c a p a c i t y i s R From\quad graph\quad ,\\ P=\frac { V }{ \sqrt { 3 } } \quad ...\quad (1)\\ and\quad \frac { dP }{ dV } \quad =\quad -\sqrt { 3 } \quad ..\quad (2)\\ Since\quad PV\quad =\quad nRT\\ PdV\quad +\quad VdP\quad =\quad nRdT\\ Now\quad using\quad 1\quad and\quad 2\quad and\quad some\quad simple\quad mainpulation,\\ PdV\quad =\quad \frac { -nRdT }{ 2 } \\ dU\quad =\quad \frac { 3 }{ 2 } nRdT\\ dq\quad =\quad dU\quad +\quad dW\\ dq\quad =\quad nRdT\\ Molar\quad heat\quad capacity\quad is\quad R

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