Chapter 9, Problem 160QP

(a)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

Explanation of Solution

The ideal gas equation states that pressure $\left(P\right)$ and volume $\left(V\right)$ of a gas is directly proportional to the temperature of the gas $\left(T\right)$ and the number of moles $\left(n\right)$ present in the gas.

$PV=nRT$ …… (1)

Here, $R$ is the universal gas constant having value $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ .

$1.25\text{ mol}$ of ${\text{N}}_2$ gas at $25°\text{C}$ and $755\text{ torr}$ is heated to $50.0°\text{C}$ and the pressure changes to $978\text{ torr}$ .

Convert temperature units from degree Celsius to kelvin units:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Convert $25°\text{C}$ from degree Celsius to kelvin units:

$\begin{array}{c}{T}_1=25+273.15\\ =\text{298}\text{.15 K}\end{array}$

Convert $50.0°\text{C}$ from degree Celsius to kelvin units:

$\begin{array}{c}{T}_2=50+273.15\\ =323.15\text{ K}\end{array}$

Covert pressure units from $\text{torr}$ to $\text{atm}$ as follows:

$\begin{array}{l}1\text{ atm}=760\text{ torr}\\ \text{1 torr}=\frac{1\text{ atm}}{760\text{ torr}}\end{array}$

Convert $755\text{ torr}$ units from $\text{torr}$ to $\text{atm}$ as follows:

$\begin{array}{c}755\text{ torr}=\frac{1\text{ atm}}{760\text{ torr}}\times 755\text{ torr}\\ =0.99\text{ atm}\end{array}$

Convert $978\text{ torr}$ units from $\text{torr}$ to $\text{atm}$ as follows:

$\begin{array}{c}978\text{ torr}=\frac{1\text{ atm}}{760\text{ torr}}\times 978\text{ torr}\\ =1.28\text{ atm}\end{array}$

Substitute $P_1$ as $0.99\text{ atm}$ , $n$ as $1.25\text{ mol}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ , and $T_1$ as $\text{298}\text{.15 K}$ in equation (1):

$\begin{array}{c}0.99\text{ atm}\times V_1=1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times \text{298}\text{.15 K}\\ V_1=\frac{1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times \text{298}\text{.15 K}}{0.99\text{ atm}}\\ =30.8\text{ L}\end{array}$

Substitute $P_2$ as $1.28\text{ atm}$ , $n$ as $1.25\text{ mol}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ , and $T_1$ as $323.15\text{ K}$ in equation (1):

$\begin{array}{c}1.28\text{ atm}\times V_2=1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 323.15\text{ K}\\ V_2=\frac{1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 323.15\text{ K}}{1.28\text{ atm}}\\ =25.8\text{ L}\end{array}$

Therefore, volume decreases as pressure increases. Hence, option A is incorrect.

The correct statement is, the volume will decrease because temperature and pressure of the gas increase.

(b)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

Explanation of Solution

Substitute $P_1$ as $0.99\text{ atm}$ , $n$ as $1.25\text{ mol}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ , and $T_1$ as $\text{298}\text{.15 K}$ in equation (1):

$\begin{array}{c}0.99\text{ atm}\times V_1=1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times \text{298}\text{.15 K}\\ V_1=\frac{1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times \text{298}\text{.15 K}}{0.99\text{ atm}}\\ =30.8\text{ L}\end{array}$

Therefore, option B is incorrect, since the initial volume of the gas is $30.8\text{ L}$ .

The correct statement is, the initial volume is $30.8\text{ L}$ .

(c)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

Explanation of Solution

Substitute $P_2$ as $1.28\text{ atm}$ , $n$ as $1.25\text{ mol}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ , and $T_1$ as $323.15\text{ K}$ in equation (1):

$\begin{array}{c}1.28\text{ atm}\times V_2=1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 323.15\text{ K}\\ V_2=\frac{1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 323.15\text{ K}}{1.28\text{ atm}}\\ =25.8\text{ L}\end{array}$

Therefore, volume decreases as temperature increase. So, option C is incorrect.

The correct statement is the volume is expected to decrease because the pressure increases by a larger factor than temperature.

(d)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

Explanation of Solution

Substitute $P_2$ as $1.28\text{ atm}$ , $n$ as $1.25\text{ mol}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ , and $T_1$ as $323.15\text{ K}$ in equation (1):

$\begin{array}{c}1.28\text{ atm}\times V_2=1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 323.15\text{ K}\\ V_2=\frac{1.25\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 323.15\text{ K}}{1.28\text{ atm}}\\ =25.8\text{ L}\end{array}$

So, the final volume is $25.8\text{ L}$ . Option D is correct.

(e)

Interpretation Introduction

Interpretation:

The correct observation is to identified.

Explanation of Solution

The initial volume is $30.8\text{ L}$ and the final volume is $25.8\text{ L}$ . Therefore, option E is incorrect.

The correct statement is, the volume of $1\text{ mol}$ of a gas at STP occupies $22.414\text{ L}$ .