The volume of an ideal gas increases 8 times and temperature becomes (1/4)th of initial temperature during a reversible change. If there is no exchange of heat in this process (ΔQ = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

The volume of an ideal gas increases 8 times and temperature becomes (1/4)th of initial temperature during a reversible change.

Q. The volume of an ideal gas increases 8 times and temperature becomes $(1/4)^{\text{th}}$ of initial temperature during a reversible change. If there is no exchange of heat in this process $(\Delta Q = 0)$ then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

A. NH₃

B. O₂

C. CO₂

D. He

Correct Answer: He

Explanation

Since there is no exchange of heat $(\Delta Q = 0)$ and the process is reversible, the process is an adiabatic process. This concept is extremely important for JEE Main, JEE Advanced and IIT JEE.

For an adiabatic process of an ideal gas:

$$ TV^{\gamma - 1} = \text{constant} $$

Given:

$$ \frac{V_2}{V_1} = 8 $$

$$ \frac{T_2}{T_1} = \frac{1}{4} $$

Using adiabatic relation:

$$ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} $$

$$ \frac{1}{4} = \left(\frac{1}{8}\right)^{\gamma - 1} $$

Writing in powers of 2:

$$ \frac{1}{4} = \left(\frac{1}{2^3}\right)^{\gamma - 1} = 2^{-3(\gamma - 1)} $$

$$ 2^{-2} = 2^{-3(\gamma - 1)} $$

Comparing powers:

$$ 3(\gamma - 1) = 2 $$

$$ \gamma = \frac{5}{3} $$

The value $\gamma = \frac{5}{3}$ corresponds to a monoatomic gas.

Among the given options, Helium (He) is a monoatomic gas.

Therefore, the correct answer is:

Explanation

Related JEE Main Questions