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

Concept:

• According to the first law of thermodynamics,

\(dq = du + dw\). where, \(dq,\; du\; \)and \(dw\) are the amount of heat change, internal energy change and work done by the system.

• When the expansion of the gas is carried out under isothermal conditions, the temperature of the system remains constant (dT=0). For a perfect gas, the pressure (P) of the system changes with the volume (V) according to the equation

PV = K, where K is a constant. 

• When the expansion of the gas is carried out under adiabatic conditions, no heat change takes place between the system and surroundings (dq=0). For a perfect gas, the pressure (P) of the system changes with the volume (V) according to the equation

​\(P{V^\gamma } = K\), where K is a constant.

• For a monoatomic gas, the value of \(\gamma \) is

​\({5 \over 3}\) (\(\gamma = {{{C_p}} \over {{C_v}}}\)).   

Explanation:

• For an adiabatic process, the pressure (P) of the system changes with the volume (V) according to the equation 

​\(P{V^\gamma } = K\)

• Under adiabatic conditions, the pressure will change with volume as, 

​\(P = {K \over {{V^\gamma }}} \) 

or, \(P \propto {1 \over {{V^\gamma }}} \)

• Now, for or a monoatomic gas, the value of \(\gamma \) is 5/3. 

So, \(P \propto {1 \over {{V^{{5 \over 3}}}}}\) 

• Thus, the pressure of the gas will fall more rapidly under adiabatic conditions because 

 \(P \propto {1 \over {{V^{{5 \over 3}}}}}\)

​Conclusion:

Hence, the pressure of the gas will fall more rapidly under adiabatic conditions because  p ∝ \(\rm \frac{1}{v^{5/3}}\)