Chemistry, 11.11.2020 14:00 ashtonbillups

A given mass of gas in a rigid container is heated from 100 °C to 300 °C. Which of the following best describes what will happen to the pressure of the gas? The pressure will

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Answer from: Carly3393

Taking into account the Gay-Lussac's Law, the pressure will increase by a factor less than three.

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

In summary, when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases.

Mathematically, Gay-Lussac's law indicates that the ratio of pressure to temperature is constant:

$\frac{P}{T} =k$

Studying two different states, an initial state 1 and a final state 2, it is satisfied:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case, a given mass of gas in a rigid container is heated from 100 ° C to 300 ° C, that is, it is heated from 373 ° K to 573 ° K. This indicates that the temperature is increasing by a factor less than three. Since pressure is directly proportional to temperature, pressure must increase in the same proportion.

Finally, the pressure will increase by a factor less than three.

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Answer from: deanlmartin

Pressure will increase by a factor of 3.

Explanation:

PV = nRT —> Ideal gas equation.

Answer from: Quest

answer: the primary difference between a hypothesis and a theory is that b hypothesis is falsifiable, but a theory has been proven to be correct. theory is an explanation of something that is happening but hypothesis is a guess of something that can happen

explanation:

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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