Q5.19P - Chemistry: Molecular Nature Of Matter And Change

Question

Each of the following processes caused the gas volume to double, as shown. For each process, state how the remaining gas variable changed or that it remained fixed:

(a) T doubles at fixed P.

(b) T and n are fixed.

(c) At fixed T, the reaction is $\mathit{C}{\mathit{D}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathit{C}\left(g\right)\mathbf{+}{\mathit{D}}_{\mathbf{2}}\left(g\right)$

(d) At fixed P, the reaction is ${\mathit{A}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{B}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{A}\mathit{B}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Step-by-Step Solution

Answer

Answer

a. At constant pressure and moles of gas, the volume of the gas will be  doubled with the doubling of temperature.

b. At constant temperature and number of moles, the volume of the gas is doubled when the pressure is decreased by half.

c. At a constant temperature, the volume of the gas is doubled when the pressure remains the same and the number of moles is doubled.

d. Thus, at constant pressure and number of moles, the volume of the gas is doubled with doubled temperature.

Step 1: Ideal gas law

The ideal gas law defines the relationship between four variables of the gas, temperature, pressure, volume, and number of moles.

The ideal gas law equation is,

$\mathit{P}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

Here,

P is pressure,

V is volume,

n is the number of moles,

R is the gas constant, and

T is the temperature.

Step 2: Subpart (a) Explanation of T doubles at fixed P.

According to ideal gas law, if T is doubles and P is fixed, then the new volume will be,

$$\begin{gathered} \mathit{P}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T} \\ \mathit{P}\mathit{V}\mathbf{\text{'}}\mathbf{=}\mathit{n}\mathit{R}\left(2T\right) \end{gathered}$$

Now, divide both equations,

$$\begin{gathered} \frac{\mathbf{PV}\mathbf{\text{'}}}{\mathbf{PV}}\mathbf{=}\frac{\mathbf{nR}\mathbf{\left(}\mathbf{2}\mathbf{T}\mathbf{\right)}}{\mathbf{nRT}} \\ \mathit{V}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathit{V} \end{gathered}$$

Thus, at constant pressure and moles of gas, the volume of the gas is doubled with the doubling of temperature.

Step 3: Subpart (b) Explanation of T and n are fixed.

According to ideal gas law, at constant temperature and moles of the gas and the volume is double with new pressure.

$\mathit{P}\mathbf{\text{'}}\left(2V\right)\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

Now divide the equation by the ideal gas equation,

$$\begin{gathered} \frac{\mathbf{P}\mathbf{\text{'}}\mathbf{\left(}\mathbf{2}\mathbf{V}\mathbf{\right)}}{\mathbf{PV}}\mathbf{=}\frac{\mathbf{nRT}}{\mathbf{nRT}} \\ \mathit{P}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{P}}{\mathbf{2}} \end{gathered}$$

Thus, at constant temperature and number of moles, the volume of the gas is doubled when the pressure is decreased by half.

Step 4: Subpart (c) At fixed T, the reaction is C D 2 ( g ) → C ( g ) + D 2 ( g ) .

The given reaction is,

$\mathit{C}{\mathit{D}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathit{C}\left(g\right)\mathbf{+}{\mathit{D}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathit{D}}_{\mathbf{2}}\left(g\right)$

Here, the number of moles of the gas is increased from 1 to 2.

According to ideal gas law, at constant temperature and the volume and number of moles are doubled, then new pressure is,

$\mathit{P}\mathbf{\text{'}}\mathbf{\left(}\mathbf{2}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathit{n}\mathit{R}\mathit{T}$

Now divide the equation by the ideal gas equation,

$$\begin{gathered} \frac{\mathbf{P}\mathbf{\text{'}}\mathbf{\left(}\mathbf{2}\mathbf{V}\mathbf{\right)}}{\mathbf{PV}}\mathbf{=}\frac{\mathbf{2}\mathbf{nRT}}{\mathbf{nRT}} \\ \mathit{P}\mathbf{\text{'}}\mathbf{=}\mathit{P} \end{gathered}$$

Now divide both equations,

$$\begin{gathered} \frac{{\displaystyle \frac{\mathbf{P}}{\mathbf{4}}\mathbf{V}\mathbf{\text{'}}}}{\mathbf{PV}}\mathbf{=}\frac{\mathbf{2}\mathbf{nRT}}{\mathbf{nRT}} \\ \mathit{V}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathit{V} \end{gathered}$$

Thus, at a constant temperature, the volume of the gas is doubled when the number of moles is doubled keeping the pressure same.

Step 5: Subpart (d) At fixed P, the reaction is A 2 ( g ) + B 2 ( g ) → 2 A B ( g ) .

The given reaction is,

${\mathit{A}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathit{B}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathit{A}\mathit{B}\left(g\right)$

As per the reaction, the number of moles of gas is the same in both reactant and product.

According to ideal gas law, at constant pressure and moles of the gas and the volume is doubled, then the new temperature is,

$\mathit{P}\mathbf{\left(}\mathbf{2}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}\mathbf{\text{'}}$

Divide the equation by the ideal gas law equation,

$$\begin{gathered} \frac{\mathbf{P}\mathbf{\left(}\mathbf{2}\mathbf{V}\mathbf{\right)}}{\mathbf{PV}}\mathbf{=}\frac{\mathbf{nRT}\mathbf{\text{'}}}{\mathbf{nRT}} \\ \mathit{T}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathit{T} \end{gathered}$$

Thus, at constant pressure and number of moles, the volume of the gas is doubled, when the temperature gets doubled.