A rigid diatomic ideal gas undergoes an adiabatic process at room temperature .The relation between temperature and volume for this process in `TV^x`= constant ,then x is :
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The correct Answer is:
D
For adiabatic process `TV^(gamma-1)`= constant So, here `x=gamma-1=7/5-1=2/5`
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VMC MODULES ENGLISH-JEE Main Revision Test-6 | JEE-2020 -PHYSICS
