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3 years ago

What happens to the average kinetic energy of the particles in a sample of matter as the temperature of the sample is increased?

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

7 0

Answer:

As the temperature of a sample of matter is increased, the average kinetic energy of the particles in the sample increase.

Explanation:

The temperature of a substance is the measure of the average kinetic energy of its partilces.

The temperature, i.e. how hot or cold is a substance, is the result of the collisions of the particles (atoms or molecules) of matter.

The kinetic theory of gases states that, if the temperature is the same, the average kinetic energy of any gas is the same, regardless the gas and other conditions.

This equation expresses it:

Avg KE = (3/2) (R / N) T

Where Avg KE is the average kinetic energy, R is the universal constant of gases, N is Avogadro's constnat, and T is the temperature measure in absolute scale (Kelvin).

As you see, in that equation Avg KE is propotional to T, which means that as the temperature is increased, the average kinetic energy increases.

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the reaction Br2(l)->Br2(g) has a deltaH=30.91 kJ and a deltaS=.0933 kJ/K. at what temperature does this reaction become spon

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Answer:

The temperature above which the reaction be spontaneous is 331.3 K.

Below this temperature; the backward reaction is the favored reaction.

Explanation:

We have an important relation from the third law of thermodynamics:

ΔG = ΔH - TΔS

ΔG is the free energy change of the reaction,

ΔH is the enthalpy change of the reaction,

ΔS is the entorpy change of the reaction,

The reaction is spontaneous when ΔG is negative.

and ΔG be negative when the value of (TΔS) is higher than the value of (ΔH).

When ΔG = 0, ΔH = TΔS.

For this reaction, ΔH = 30.91 KJ and ΔS = 0.0933 KJ/K.

The temperature above which the reaction be spontaneous is:

T = ΔH / ΔS = (30.91 KJ) / (0.0933 KJ/K) = 331.29 K ≅ 331.3 K.

∴ The temperature above which the reaction be spontaneous is 331.3 K.

Below this temperature; the backward reaction is the favored reaction.

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if an unknown gas has one third the root mean squarevspeed of H2 at 300 K, what is the molar mass of the gas?

Viefleur [7K]

The molar mass of the gas : 18 x 10⁻³ kg/mol

<h3>Further explanation</h3>

Given

An unknown gas has one third the root mean square speed of H2 at 300 K

Required

the molar mass of the gas

Solution

Average velocities of gases can be expressed as root-mean-square (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

T = temperature, Mm = molar mass of the gas particles , kg/mol

R = gas constant 8,314 J / mol K

v rms An unknown gas = 1/3 v rms H₂

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$\tt v_{rms}=\sqrt{\dfrac{3\times 8.314\times 300}{2.10^{-3}} }\\\\v_{rms}=1934.22$

V rms of unknown gas =

$\tt \dfrac{1}{3}\times 1934.22=644.74$

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