Question

The volume of an ideal gas increases \( 8 \) times and the temperature becomes \( \frac{1}{4} \) of the 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. He
• B. O$_2$
• C. CO$_2$
• D. NH$_3$

Hint:

For reversible adiabatic processes, use $TV^{\gamma-1}=\text{constant}$. A value $\gamma=\frac{5}{3}$ always indicates a monoatomic ideal gas.

Answer 1

The Correct Option is A

Since there is no exchange of heat $(\Delta Q=0)$ and the process is reversible, the process is adiabatic.
Step 1: Use the adiabatic relation between temperature and volume.
For a reversible adiabatic process of an ideal gas: \[ TV^{\gamma-1}=\text{constant} \] Step 2: Apply the given data.
Given: \[ V_2 = 8V_1,\quad T_2 = \frac{1}{4}T_1 \] Substituting into the adiabatic relation: \[ T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1} \] \[ T_1V_1^{\gamma-1} = \frac{1}{4}T_1(8V_1)^{\gamma-1} \] Canceling $T_1V_1^{\gamma-1}$: \[ 1 = \frac{1}{4}\,8^{\gamma-1} \] Step 3: Solve for $\gamma$.
\[ 8^{\gamma-1} = 4 \] Writing in powers of $2$: \[ (2^3)^{\gamma-1} = 2^2 \Rightarrow 3(\gamma-1)=2 \] \[ \gamma = \frac{5}{3} \] Step 4: Identify the gas.
The ratio of specific heats $\gamma=\frac{5}{3}$ corresponds to a monoatomic gas.
Among the given options, only Helium (He) is monoatomic.
Final Answer: $\boxed{\text{He}}$