Question 411: Thermodynamics & KTG

Question

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV
x
= constant, then x is :

Accepted Answer

For adiabatic process : TV
$$\gamma $$$$-$$1
= constant
For diatomic process : $$\gamma $$$$-$$1  = $${7 \over 5} - 1$$
$$ 	herefore $$  x = $${2 \over 5}$$

Answer

$${5 \over 3}$$

Answer

$${2 \over 5}$$

Answer

$${3 \over 5}$$

Answer

$${2 \over 3}$$

Thermodynamics & KTG

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV x = constant, then x is :

Choose the correct answer: