A monoatomic ideal gas is compressed adiabatically to $\left(\frac{1}{27}\right)$ of its initial volume. If initial temperature of the gas is ' T ' K and final temperature is ' xT ' K , the value of ' x ' is

The temperature change in an adiabatic process for an ideal gas follows $TV^{\gamma-1} = \text{constant}$. For a monatomic gas, $\gamma = \frac{5}{3}$.

Given $T_1 = T$, $T_2 = xT$, $V_1$, and $V_2 = \frac{V_1}{27}$, the adiabatic relation yields:

$T V_1^{2/3} = xT \left(\frac{V_1}{27}\right)^{2/3}$

Dividing both sides by $T$ and $V_1^{2/3}$ gives $1 = \frac{x}{27^{2/3}}$.

Since $27^{2/3} = (3^3)^{2/3} = 3^2 = 9$, the equation simplifies to $1 = \frac{x}{9}$.

The solution is therefore $x = 9$.