question_answer

Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is:

A)  30 K                           

B)  18 K   

C)       50 K               

D)       42 K

Correct Answer: D

Solution :

F-B-D of piston of cylinder A Where,\[{{P}_{0}}\]is the atmospheric pressure, mg the weight of the piston, a the acceleration of the piston and P, the pressure of the gas. Invoking Newton's Second law, \[F=ma,\]we get: \[PA-{{P}_{0}}A-mg=ma\] Assuming slow movement, the acceleration of the piston, i.e., \[a=0,\]we get: \[PA-{{P}_{0}}A-mg=0\Rightarrow P=\frac{mg+{{P}_{0}}A}{A}=\]constant So, for A, the process is Isobaric.             \[\therefore \]\[\Delta Q=\mu {{C}_{p}}(\Delta \Tau )\,t\] For B the process is Isochoric, \[\therefore \]\[\Delta Q=\mu C,{{(\Delta \Tau )}_{2}}\] Now, \[{{C}_{p}}{{(\Delta T)}_{1}}={{C}_{v}}{{(\Delta T)}_{2}}\]or \[\frac{7R}{2}\times 30=\frac{5R}{2}{{(\Delta \Tau )}_{2}}\] \[\Delta {{\Tau }_{2}}=42\,K\] Hence, the correction option is [d].