JEE Main 2019 — Physics solution

JEE Main2019Physics

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.

A
$\frac{2}{5}$
B
$\frac{2}{3}$
C
$\frac{5}{3}$
D
$\frac{3}{5}$

Solution

For adiabatic

PV^$γ$ = constant

TV^x = constant

$\Rightarrow$ PVV^x = constant

x + 1 = $γ$

x = 2/5

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