Chapter 6, Problem 6.91QE

Interpretation Introduction

Interpretation:

Pressure in each container before and after the stopcock is opened has to be calculated.

Concept Introduction:

The properties that are dependent upon the size or the amount of eth matter contained in the system are termed as extensive properties. They are additive in nature.

Any gas obeys the assumption laid down in kinetic molecular theory is said to be an ideal gas. The combination of all the gas laws namely Boyle’s law, Charles law and Avogadro’s leads to the ideal gas equation that can be used to relate all the four properties such as given below:

  $PV=nRT$        (1)

Here,

$\text{R}$ is the universal gas constant.

$V$ denotes the volume.

$n$ denotes the number of moles.

$T$ denotes temperature.

$P$ denotes the pressure

Answer to Problem 6.91QE

Pressure in the left and right container before the stopcock is opened are $0.985\text{ atm}$ and $0.493\text{ atm}$ respectively while the Pressure when the stopcock is opened is $0.739\text{ atm}$.

Explanation of Solution

Since each dot represents $0.005\text{ mol}$ and there are 8 dots hence the total number of moles of gas contained in the left container is calculated as follows:

  $\begin{array}{c}\text{Number of moles}=0.005\text{ mol}\left(8\right)\\ =0.04\text{ mol}\end{array}$

Similarly, since there are four blue molecules that represent $0.005\text{ mol}$ hence the total number of moles of gas contained in the right container is calculated as follows:

  $\begin{array}{c}\text{Number of moles}=0.005\text{ mol}\left(4\right)\\ =0.02\text{ mol}\end{array}$

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Here,

$T\left(\text{K}\right)$ denotes the temperature in kelvins.

$T\left(°\text{C}\right)$ denotes the temperature in Celsius.

Substitute $\text{27 }°\text{C}$ for $T\left(\text{K}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=27\text{ °C}+273\text{ K}\\ =300\text{ K}\end{array}$

Rearrange equation (1) to obtain the expression of temperature as follows:

  $P=\frac{nRT}{V}$        (2)

Substitute $0.04\text{ mol}$ for $n$, $\text{1 L}$ for $V$, and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ and $\text{300 K}$ for $T$ in equation (2) to calculate the pressure in the left container.

  $\begin{array}{c}P=\frac{\left(0.04\text{ mol}\right)\left(0.08206\text{ L}\text{.atm/mol}\text{.K}\right)\left(\text{300 K}\right)}{\text{1 L}}\\ =0.985\text{ atm}\end{array}$

Substitute $0.02\text{ mol}$ for $n$, $\text{1 L}$ for $V$,and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ and $\text{300 K}$ for $T$ in equation (2)to calculate the pressure in the left container.

  $\begin{array}{c}P=\frac{\left(0.02\text{ mol}\right)\left(0.08206\text{ L}\text{.atm/mol}\text{.K}\right)\left(\text{300 K}\right)}{\text{1 L}}\\ =0.493\text{ atm}\end{array}$

After the stopcock is opened the volume is taken as the sum of the volume of both containers as it is an extensive property. Thus the total volume is calculated as follows:

  $V=V_1+V_2$        (3)

Here,

$V_1$ denotes the volume of container 1.

$V_2$ denotes the volume of container 2.

Substitute $\text{1 L}$ for $V_1$ and $\text{1 L}$ for $V_2$ in equation (3).

  $\begin{array}{c}V=\text{1 L}+\text{1 L}\\ =\text{2 L}\end{array}$

Similarly, the amount of substance or the number of moles is also an additive property and therefore the total number of moles $\left(n\right)$ is calculated as follows:

  $n=n_1+n_2$        (4)

Here,

$n_1$ denotes the volume of container 1.

$n_2$ denotes the volume of container 2.

Substitute $0.04\text{ mol}$ for $n_1$, $0.02\text{ mol}$ for $n_2$ in equation (4).

  $\begin{array}{c}n=0.04\text{ mol}+0.02\text{ mol}\\ =0.06\text{ mol}\end{array}$

Therefore substitute $0.06\text{ mol}$ for $n$, $\text{2 L}$ for $V$, $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $\text{300 K}$ for $T$ in equation (2) to calculate the pressure when the stopcock is opened.

  $\begin{array}{c}P=\frac{\left(0.06\text{ mol}\right)\left(0.08206\text{ L}\text{.atm/mol}\text{.K}\right)\left(\text{300 K}\right)}{\text{2 L}}\\ =0.739\text{ atm}\end{array}$