Question 174: Thermodynamics & KTG

Question

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

Accepted Answer

It is given that there is no exchange of heat in this process, which means $\Delta \mathrm{Q}=0$, i.e., this is an adiabatic process.
For an ideal gas undergoing a reversible adiabatic process, the relationship between Temperature (T) and Volume (V) is given by the equation:
$$ \mathrm{T} \cdot \mathrm{~V}^{\gamma-1}=	ext { constant } $$
Where $\gamma$ is the adiabatic index or the ratio of specific heats $\left(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}ight)$. This $\gamma$ value is different depending atomicity of the gases (monoatomic, diatomic, polyatomic).
$$ \mathrm{T}_1 \cdot \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \cdot \mathrm{~V}_2^{\gamma-1} $$
Final volume $\left(V_2ight)$ is 8 times the initial volume $\left(V_1ight) \Rightarrow V_2=8 V_1$
Final temperature $\left(T_2ight)$ is $\left(\frac{1}{4}ight)^{	ext {th }}$ the initial temperature $\left(T_1ight) \Rightarrow T_2=\frac{T_1}{4}$
$$ \mathrm{T}_1 \cdot \mathrm{~V}_1^{\gamma-1}=\left(\frac{\mathrm{T}_1}{4}ight) \cdot\left(8 \mathrm{~V}_1ight)^{\gamma-1} $$
$$\Rightarrow $$ $\mathrm{V}_1^{\gamma-1}=\frac{1}{4} \cdot 8^{\gamma-1} \cdot \mathrm{~V}_1^{\gamma-1}$
$\Rightarrow 1=\frac{1}{4} \cdot 8^{\gamma-1} $
$\Rightarrow  4=8^{\gamma-1} $
$\Rightarrow 2^2=2^{3(\gamma-1)} $
$\Rightarrow 2=3(\gamma-1) $$ 2=3 \gamma-3 $
$\Rightarrow 5=3 \gamma $
$\Rightarrow \gamma=\frac{5}{3}$
The value of $\gamma$ is directly related to the degrees of freedom (f) of the gas molecules by the formula :
$$ \gamma=1+\frac{2}{f} $$
$\Rightarrow \frac{5}{3}=1+\frac{2}{f}$
$\Rightarrow \frac{5}{3}-1=\frac{2}{f}$
$$\Rightarrow \frac{2}{3}=\frac{2}{f}$$
$$\Rightarrow f=3$$
A gas with 3 degrees of freedom (all translational, no rotational or vibrational modes activated) is a monoatomic gas.
$\mathrm{NH}_3$ (Ammonia) is polyatomic molecule (non-linear).
$\mathrm{O}_2$ (Oxygen) is a diatomic molecule.
$\mathrm{CO}_2$ (Carbon Dioxide) is polyatomic molecule (linear).
He (Helium) is a noble gas, exists as single atoms. It is a monoatomic gas. Therefore, the given gas is a monoatomic gas (He).
Hence, the correct option is (D).

Answer

$\mathrm{NH}_3$

Answer

$\mathrm{O}_2$

Answer

$\mathrm{CO}_2$

Answer

He

Thermodynamics & KTG

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