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3 years ago

By what factor must the absolute temperature of a gas be increased to increase the rms speed of its molecules by a factor of 3?

Chemistry

1 answer:

attashe74 [19]3 years ago

6 0

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

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3 years ago

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2 years ago

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3 years ago

Read 2 more answers

In an equimolar mixture of o2 and n2, are the mass fractions equal?

nika2105 [10]

Answer:

Explanation:

The mass fraction is defined as:

$w_{i}=\frac{m_{i} }{m_{t} }$

where:

wi: mass fraction of the substance i
mi: mass of the substance i
mt: total mass of the system

The mass fraction of two substances (A and B), will be the same, ONLY if the mass of the substance A (mA) is the same as the mass of the substance B (mB).

An equimolar mixutre of O2 and N2 has the same amount of moles of oxygen and nitrogen, just to give an example let's say that the system has 1 mole of O2 and 1 mole of N2. Then using the molecuar weigth of each of them we can calculate the mass:

mA= 1 mole of O2 * 16 g/1mol = 16 g

mB=1 mole of N2 *28 g/1mol=28 g

As mA≠mB then the mass fractions are not equal, so the answear is NO.

4 0

3 years ago

2.2 Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu; 83.79% of 52Cr, with an

shutvik [7]

Answer:

Average atomic mass = 51.9963 amu

Explanation:

Given data:

Abundance of Cr⁵⁰ with atomic mass= 4.34% , 49.9460 amu

Abundance of Cr⁵² with atomic mass = 83.79%, 51.9405 amu

Abundance of Cr⁵³ with atomic mass =9.50%, 52.9407 amu

Abundance of Cr⁵⁴ with atomic mass = 2.37%, 53.9389 amu

Average atomic mass = 51.9963 amu

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n) / 100

Average atomic mass = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100

Average atomic mass = 216.7656 + 4352.0945 + 502.9367 +127.8352 / 100

Average atomic mass = 5199.632 / 100

Average atomic mass = 51.9963 amu

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3 years ago