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The temperature of low-pressure hydrogen is reduced from 100 °C

Berikut ini adalah pertanyaan dari muhanmadwahyu pada mata pelajaran Fisika untuk jenjang Sekolah Menengah Atas

The temperature of low-pressure hydrogen is reduced from 100 °C to 20 °C. The rms speed of its molecules decreases by approximatelya. 80%
b. 89%
c. 46%
d. 21%
e. 11%

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

The RMS speed of its molecules decreases by approximately 11%. E is the answer. The RMS speed after the temperature decrease is 89% RMS speed before the temperature decrease.

Penjelasan dengan langkah-langkah:

Diketahui:

T₁ = 100 °C = 100 + 273 = 373 K
T₂ = 80 °C = 80 + 273 = 293 K

Ditanyakan:

RMS speed reduction?

Jawaban:

RMS speed is the square root of the mean of the square of speeds of all molecules of gas.

$v_{RMS} \:=\: \sqrt{\frac{3kT}{m}}$

k = Boltzmann constant = 1.38 × 10⁻²³ m²kg/s²K
T = temperature gas (K)
m = the mass of one gas molecule

If there is only a change in system temperature then the RMS speed can be compared with the temperature because m is the same and k never changes.

$\frac{v_{RMS2}}{v_{RMS1}} \:=\: \sqrt{\frac{T_2}{T_1}}$

$\frac{v_{RMS2}}{v_{RMS1}} \:=\: \sqrt{\frac{293}{373}}$

$\frac{v_{RMS2}}{v_{RMS1}} \:=\: \sqrt{0.7855}$

$\frac{v_{RMS2}}{v_{RMS1}} \:=\: 0.8863$

$v_{RMS2} \:=\: 0.8863 \: v_{RMS1}$

$v_{RMS2} \:=\: 0.8863 \: v_{RMS1} \times 100\%$

$v_{RMS2} \:=\: 88.63\% \: v_{RMS1}$

$v_{RMS2} \:=\: 89\% \: v_{RMS1}$

RMS speed reduction

= $v_{RMS1} \:-\: v_{RMS2}$

= $100\% \: v_{RMS1} \:-\: 89\% \: v_{RMS1}$

= $11\% \: v_{RMS1}$

E

Pelajari lebih lanjut

Materi tentang Kecepatan RMS yomemimo.com/tugas/20970517

#BelajarBersamaBrainly #SPJ1

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Last Update: Fri, 31 Mar 23

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