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One mole of a mono- atomic ideal gas und...

One mole of a mono- atomic ideal gas undergoes a quasi- static process, which is depicted by a straight line joining points `(V_(0)T_(0)) " and " (2V_(0) , 3T_(0))` in a V - T diagram . What is the value of the heat capacity of the gas at the point `(V_(0) , T_(0))` ?

A

R

B

`3/2 R`

C

`2 R`

D

0

Text Solution

Verified by Experts

The correct Answer is:
c
Since path is isochoric .
` :. ` Work done , `dW = PdV` Change in internal energy is, `Delta U = nC_(V) Delta T`
where `C_(V)` = specific heat at constant volume.
Heat given in process is , `Delta Q = nCDelta T `,

where n is number of moles of gas , C is molar heat capacity.
According to graph, temperature is increasing with increase in volume. It means that energy is used both in increasing internal energy and work done.
According to `1^(st)` law of thermodynamics
` PdV + nC_(V)dT = nCdT`
or ` P/n (dV)/(dT) + C_(V) = C` .....(i)
At `(V_(0) , T_(0)) (dV)/(dT) = V_(0)/(2T_(0)) ` ....(ii)
Again , `P_(0)V_(0) = nRT_(0) rArr P_(0)/n = (RT_(0))/V_(0) ` ....(iii)
From (i), (ii) and (iii) ,
`C = (RT_(0))/V_(0) V_(0)/(2T_(0)) + 3/2 R = 2 R [ :. " for monoatomic gas , " C_(V) = 3/2 R]`
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