Respuesta :
Answer:
The temperature of the molecule will be increased by a factor of 9
Explanation:
The relationship between the temperature and the rms speed of a gas is given by
Vrms=√(3RT/MM)
Where R = Universal gas constant
T = Absolute temperature of the gas in kelvin
MM = Molar mass of the gas
Since from the question, only the rms speed and the temperature are said to change, the term 3R/MM can be represented by a constant k. The relationship then becomes Vrms=√kT
At the intitial speed, temperature = T1
Vrms=√(kT1 ) (1)
When the speed is increased by a factor of 3, the relationship becomes
(3V)rms=√(kT2 ) (2)
Dividing equation (2) by equation (1)
(3V)rms/Vrms =√(kT2 )/√(kT1 )
3=√(T2/T1 )
T2/T1 =9
The temperature of the molecule will be increased by a factor of 9
The temperature of the molecule will be increased by a factor of 9
The relationship between the temperature and the rms speed of a gas is given by
[tex]V_{rms}=\sqrt{(3RT/M)}[/tex]
where R = Universal gas constant
T = Absolute temperature of the gas in kelvin
M = Molar mass of the gas
Since from the question, only the rms speed and the temperature are said to change, the term 3R/M can be represented by a constant k. The relationship then becomes
[tex]V_{rms}=\sqrt{kT }[/tex]
At the initial speed, temperature = T₁
[tex]V_{rms}=\sqrt{(kT_1 ) }[/tex] ............(i)
When the speed is increased by a factor of 3, the relationship becomes
[tex](3)V_{rms}=\sqrt{(kT_2)}[/tex]..................(ii)
Dividing equation (ii) by equation (i)
[tex]\frac{ (3)V_{rms}}{V_{rms}} =\frac{\sqrt{(kT_2)} }{\sqrt{(kT_1 )} }\\\\3=\frac{\sqrt{(T_2)} }{\sqrt{(T_1 )} }\\\\\frac{T_2}{T_1} =9[/tex]
The temperature of the molecule will be increased by a factor of 9
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brainly.com/question/5343098
