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

Given the problem, we need to determine which gas behaves in a manner consistent with the changes in volume and temperature given, under the condition of no exchange of heat (\(\Delta Q = 0\)) in a reversible process.

This condition suggests that the process is adiabatic, characterized by the equation:

\(PV^{\gamma} = \text{constant}\)

where \(\gamma\) is the adiabatic index or heat capacity ratio, defined as \(\frac{C_p}{C_v}\).

Also, for an ideal gas, the equation relating volume and temperature in an adiabatic process is:

\(TV^{\gamma-1} = \text{constant}\)

Given in the problem:

• Volume increases 8 times: \(V_2 = 8V_1\)
• Temperature decreases to \(\frac{1}{4}\) of its initial value: \(T_2 = \frac{T_1}{4}\)

Substituting into the adiabatic equation gives us:

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

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

Taking the logarithm on both sides, we get:

\(-\log(4) = (\gamma-1)(-\log(8))\)

Simplifying:

\(\log(4) = \log(8)(\gamma-1)\)

\(\Rightarrow \gamma-1 = \frac{\log(4)}{\log(8)}\)

The relationship \(2^2 = 4\) and \(2^3 = 8\) can be used:

\(\Rightarrow \gamma-1 = \frac{2 \log(2)}{3 \log(2)}\)

\(\Rightarrow \gamma-1 = \frac{2}{3}\)

\(\Rightarrow \gamma = \frac{5}{3}\)

The adiabatic index \(\gamma = \frac{5}{3}\) is characteristic of a monoatomic ideal gas, like helium (He).

Thus, the correct answer, based on these calculations and the condition given, is that the gas is helium (He).