Question 768: Thermodynamics & KTG

Question

An ideal gas $(\gamma=1.5)$ is expanded adiabatically.
To reduce root mean square velocity of molecules two times, the gas should be expanded

Accepted Answer

To understand how the expansion of an ideal gas affects the root mean square (r.m.s) velocity of its molecules during an adiabatic process, let's break down the steps:
Relationship of r.m.s. Velocity and Temperature
:
The root mean square velocity ($v$) of gas molecules is proportional to the square root of its absolute temperature ($T$), given by:
$    v \propto \sqrt{T}    $
Therefore, the ratio of the r.m.s velocities under two different temperatures is:
$    \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \quad 	ext{(Equation 1)}    $
Condition for Reduced r.m.s. Velocity
:
If the r.m.s. velocity is reduced by half, then:
$    v_2 = \frac{v_1}{2}    $
Substituting this into Equation 1 gives:
$    \frac{1}{2} = \sqrt{\frac{T_2}{T_1}} \quad \Rightarrow \quad \frac{T_1}{T_2} = 4 \quad 	ext{(Equation 2)}    $
Adiabatic Expansion
:
For an adiabatic process, which conserves internal energy without heat exchange, the relationship is:
$    T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}    $
Substituting the temperature ratio from Equation 2, we have:
$    \left(\frac{V_2}{V_1}ight)^{\gamma-1} = \frac{T_1}{T_2} = 4    $
Solving for Volume Expansion
:
Using $\gamma = 1.5$, the equation becomes:
$    \left(\frac{V_2}{V_1}ight)^{0.5} = 4    $
Solving for $\frac{V_2}{V_1}$, we square both sides:
$    \frac{V_2}{V_1} = 16    $
Thus, the gas should be expanded 16 times to reduce the r.m.s. velocity of its molecules to half its initial value through adiabatic expansion.

Answer

20 times

Answer

16 times

Answer

12 times

Answer

8 times

Thermodynamics & KTG

An ideal gas (\gamma=1.5) is expanded adiabatically. To reduce root mean square velocity of molecules two times, the gas should be expanded

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