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One mole of a monoatomic ideal gas under...

One mole of a monoatomic ideal gas undergoes the process `ArarrB` in the given P-V diagram. The specific heat for this process is

A

`(21R)/(10)`

B

`(18R)/(10)`

C

`(9R)/10`

D

`(13R)/10`

Text Solution

Verified by Experts

The correct Answer is:
A

`Delta Q_(A to B ) = Delta U_(A to B) + W_(A to B)`
`r/2 nR(T_B - T_B)+1/2 (P_0 + 2P0-) (3V_0 - V_0) = 3/2 (P_B V_B - P_A V_A ) + 3 P_0 V_0 = 15/2 P_0 V_0 + 3 P_0 V_0 = 21/ 2 P_0 V_0`
`DeltaT_(A to B) = ((P_B V_B - P_A V_A))/R = (5 P_0 V_0)/R " "rArr" av. Molar specific heat " = (Delta Q_(A to B))/(Delta T_(A to B)) = 21/10 R`
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