A monatomic perfect gas undergoes expansion from (p₁, V₁) to (P₂, V₂) under isothermal or adiabatic conditions. The pressure of the gas will fall more rapidly under adiabatic conditions because

Question

A monatomic perfect gas undergoes expansion from (p1, V1) to (P2, V2) under isothermal or adiabatic conditions. The pressure of the gas will fall more rapidly under adiabatic conditions because

Prepp Team · Accepted Answer

Understanding Monatomic Gas Expansion

We are looking at the expansion of a monatomic perfect gas under two different conditions: isothermal and adiabatic. We want to understand why the pressure falls more rapidly during adiabatic expansion compared to isothermal expansion when the volume increases from $V_1$ to $V_2$.

The rate at which pressure falls during expansion depends on the relationship between pressure ($p$) and volume ($V$) for the specific process.

Isothermal Expansion Condition

During an isothermal process, the temperature ($T$) of the gas remains constant. For a perfect gas, the equation of state is $pV = nRT$. Since $n$ (number of moles) and $R$ (gas constant) are constant, and $T$ is constant for an isothermal process, the product $pV$ is constant.

$pV = \text{constant}$

This implies that pressure is inversely proportional to volume:

$p \propto \frac{1}{V}$

We can write this as $p \propto \frac{1}{V^1}$. In this case, the pressure decreases as volume increases, and the decrease rate is related to $1/V^1$.

Adiabatic Expansion Condition

During an adiabatic process, no heat is exchanged between the gas and its surroundings ($Q=0$). For a perfect gas, the relationship between pressure ($p$) and volume ($V$) is given by:

$pV^\gamma = \text{constant}$

Here, $\gamma$ (gamma) is the adiabatic index, which is the ratio of the molar specific heat capacity at constant pressure ($C_p$) to the molar specific heat capacity at constant volume ($C_v$), i.e., $\gamma = C_p/C_v$.

This equation shows that pressure is inversely proportional to volume raised to the power $\gamma$:

$p \propto \frac{1}{V^\gamma}$

Adiabatic Index ($\gamma$) for Monatomic Gas

For a monatomic perfect gas, the degrees of freedom are 3 (translational only). According to the kinetic theory of gases, the molar specific heat capacity at constant volume ($C_v$) for a monatomic gas is $\frac{3}{2}R$. Using Mayer's relation ($C_p - C_v = R$), the molar specific heat capacity at constant pressure ($C_p$) is $C_p = C_v + R = \frac{3}{2}R + R = \frac{5}{2}R$.

The adiabatic index $\gamma$ for a monatomic perfect gas is therefore:

$\gamma = \frac{C_p}{C_v} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}$

So, for a monatomic perfect gas undergoing adiabatic expansion, the relationship between pressure and volume is:

$pV^{5/3} = \text{constant}$

Or

$p \propto \frac{1}{V^{5/3}}$

Comparing Pressure Fall Rates During Expansion

During expansion, the volume $V$ increases. Let's compare how pressure changes with volume in the two processes:

Isothermal expansion: $p \propto \frac{1}{V^1}$
Adiabatic expansion (monatomic gas): $p \propto \frac{1}{V^{5/3}}$

To see which pressure falls faster, we compare the exponents of $V$. The exponent for isothermal expansion is 1. The exponent for adiabatic expansion of a monatomic gas is 5/3.

Since $\frac{5}{3} = 1.66\dots$ and $1.66\dots > 1$, the pressure decreases more steeply with increasing volume when $p$ is proportional to $1/V^{5/3}$ than when $p$ is proportional to $1/V^1$.

Therefore, the pressure of the gas falls more rapidly under adiabatic conditions for a monatomic gas because pressure is inversely proportional to $V^{5/3}$, and the exponent 5/3 is greater than the exponent 1 in the isothermal case.

The relationship that describes the pressure variation during adiabatic expansion of a monatomic perfect gas is $p \propto \frac{1}{V^{5/3}}$.