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For a thermodynamic process specific heat of gas is zero . The process is 

Option: 1

isothermal 


Option: 2

isochoric 


Option: 3

isobaric 


Option: 4

Adiabatic 


Answers (1)

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As we have learned

Specific heat for adiabatic process -

C= \frac{\Delta Q}{n\Delta T}= 0

 

- wherein

since in the adiabatic process \Delta Q=0

 

 Specific heat = \frac{\Delta Q}{m\Delta T} \\\\ \Delta Q = 0 \: \: for \: \: adiabatic \: \: process \\\\ C = \frac{0 }{m \Delta T} = 0

 

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