Adiabatic Process Calculation

The process described is a reversible change with no heat exchange ($\Delta Q = 0$). This signifies an adiabatic process for the ideal gas.

Adiabatic Process Formula

The relationship between temperature ($T$) and volume ($V$) during an adiabatic process for an ideal gas is:

$ T V^{\gamma-1} = \text{constant} $

Here, $\gamma$ represents the adiabatic index, which depends on the gas type.

Applying Given Conditions

Let the initial state be (1) and the final state be (2). The formula can be written as:

$ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} $

The problem states:

• Volume increases 8 times: $V_2 = 8V_1$
• Temperature becomes 1/4th: $T_2 = \frac{1}{4} T_1$

Calculating the Adiabatic Index ($\gamma$)

Substitute the given values into the adiabatic equation:

$ T_1 V_1^{\gamma-1} = \left(\frac{1}{4} T_1\right) (8V_1)^{\gamma-1} $

Divide both sides by $T_1$:

$ V_1^{\gamma-1} = \frac{1}{4} (8V_1)^{\gamma-1} $

Rearrange to isolate the volume terms:

$ 4 = \frac{(8V_1)^{\gamma-1}}{V_1^{\gamma-1}} $

Simplify using exponent rules:

$ 4 = \left(\frac{8V_1}{V_1}\right)^{\gamma-1} $

$ 4 = 8^{\gamma-1} $

Express both sides with the base 2:

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

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

Equate the exponents:

$ 2 = 3(\gamma-1) $

Solve for $\gamma$:

$ \gamma-1 = \frac{2}{3} $

$ \gamma = 1 + \frac{2}{3} = \frac{5}{3} $

Identifying the Gas

The calculated adiabatic index is $\gamma = 5/3$. This value is characteristic of monatomic ideal gases.

From the given options, He (Helium) is a monatomic gas.