Respuesta :
Answer:
Given that
V2/V1= 0.25
And we know that in adiabatic process
TV^န-1= constant
So
T1/T2=( V1 /V2)^ န-1
So = ( 1/0.25)^ 0.66= 2.5
Also PV^န= constant
So P1/P2= (V2/V1)^န
= (1/0.25)^1.66 = 9.98
A. RMS speed is
Vrms= √ 3RT/M
But this is also
Vrms 2/Vrms1= (√T2/T1)
Vrms2=√2.5= 1.6vrms1
B.
Lambda=V/4π√2πr²N
So
Lambda 2/lambda 1= V2/V1 = 0.25
So the mean free path can be inferred to be 0.25 times the first mean free path
C. Using
Eth= 3/2KT
So Eth2/Eth1= T2/T1
So
Eth2= 2.5Eth1
D.
Using CV= 3/2R
Cvf= Cvi
So molar specific heat constant does not change
The rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.
Given that Â
[tex]\bold{\dfrac {V_2}{V_1}= 0.25}[/tex]
In adiabatic process Â
[tex]\bold {TV^\gamma^-^1}[/tex]= constant Â
So Â
[tex]\bold {\dfrac {T1}{T2}=(\dfrac {V1 }{V2})^ \gamma^-^1}}\\[/tex] Â
So,
[tex]\bold {\dfrac {T_1} {T_2} = (\dfrac { 1}{0.25})^ 0^.^6^6= 2.5}[/tex] Â
for pressure,
[tex]\bold {PV^\gamma = \ Constant }[/tex] Â
So
[tex]\bold {\dfrac {P1}{P2}=(\dfrac {V1 }{V2})^ \gamma}}\\\\\bold {\dfrac {P1}{P2}=(\dfrac {1}{0.25})^ 0^.^6^6} = 9.98 } }[/tex] Â
A. rms speed can be calculated as
Â
[tex]\bold {\dfrac {Vrms 2}{Vrms1}= \sqrt {T2T1})}\\\\\bold {Vrms2 =\sqrt {2.5} = 1.6\ Vrms1 }[/tex]
B. Â The mean free path can be calculated as
[tex]\bold {\dfrac {\lambda_1 }{\lambda_2} = \dfrac {V_1}{V_2} = 0.25 \\ }[/tex]
Mean free path is 0.25 times the first mean free path.
C.
[tex]\bold {Eth= \dfrac {3}{2}kT}}\\\\\bold {\dfrac {E_t2}{E_t1} = \dfrac {T_2}{T_1}}\\\\\bold {E_t2= 2.5\ E_t1}[/tex] Â
Â
D. Â the molar specific at a constant volume can be calculated a using,
[tex]\bold {CV= \dfrac 3{2}R }[/tex] Â
 [tex]\bold {Cv_f= Cv_i }[/tex]
So, molar specific heat constant will not change.
Therefore, the rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.
To know more about adiabatic processes,
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