JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 27)
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = constant, then x is :
$${5 \over 3}$$
$${2 \over 5}$$
$${3 \over 5}$$
$${2 \over 3}$$
Forklaring
For adiabatic process : TV$$\gamma $$$$-$$1 = constant
For diatomic process : $$\gamma $$$$-$$1 = $${7 \over 5} - 1$$
$$ \therefore $$ x = $${2 \over 5}$$
For diatomic process : $$\gamma $$$$-$$1 = $${7 \over 5} - 1$$
$$ \therefore $$ x = $${2 \over 5}$$