Chapter 7, Problem 7.36EP

(a)

Interpretation Introduction

Interpretation:

The pressure in millimeter of mercury for a sample of carbon monoxide gas at given set of conditions by using combined gas law has to be determined.

Concept Introduction:

Combined gas law tells that the Kelvin temperature of gas is directly proportional to the pressure and volume of a fixed amount of gas.  The mathematical expression for combined gas law can be represented as follows,

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Explanation of Solution

Record the given data,

$\begin{array}{l}{P}_1=735mmHg\\ V_1=\text{7}\text{.31}L\\ T_1=45°Cconvertedto318K\\ P_2=?\\ T_2=357°Cconvertedto630K\\ V_2=13.5L\end{array}$

Now, substitute these values in rearranged combined gas law and do some simple mathematical calculation to gent final answer as follows,

$P_2=P_1\times \frac{V_1}{V_2}\times \frac{T_2}{T_1}$

$P_2=735mmHg\times \left(\frac{7.31L}{13.5L}\right)\times \left(\frac{630K}{318K}\right)=788mmHg$

The final pressure in millimeter of mercury for a sample of carbon monoxide gas at given set of conditions is $788mmHg.$

(b)

Interpretation Introduction

Interpretation:

The temperature in degree Celsius for a sample of carbon monoxide gas at given set of conditions by using combined gas law has to be determined.

Concept Introduction:

Combined gas law tells that the Kelvin temperature of gas is directly proportional to the pressure and volume of a fixed amount of gas.  The mathematical expression for combined gas law can be represented as follows,

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Explanation of Solution

Record the given data,

$\begin{array}{l}{P}_1=735mmHg\\ V_1=\text{7}\text{.31}L\\ T_1=45°Cconvertedto318K\\ P_2=1275mmHg\\ T_2=?\\ V_2=0.800L\end{array}$

Now, substitute these values in rearranged combined gas law and do some simple mathematical calculation to gent final answer as follows,

$T_2=T_1\times \frac{V_2}{V_1}\times \frac{P_2}{P_1}$

$T_2=318K\times \left(\frac{1275mmHg}{735mmHg}\right)\times \left(\frac{\text{0}\text{.800L}}{\text{7}\text{.31L}}\right)=60K$

The obtained pressure is in Kelvin units now we have to convert this value into degree Celsius as follows,

$T-273$

$60-273=-213°C$

The final temperature in degree Celsius for a sample of carbon monoxide gas at given set of conditions is $-213°C.$

(c)

Interpretation Introduction

Interpretation:

The volume in liters for a sample of carbon monoxide gas at given set of conditions by using combined gas law has to be determined.

Concept Introduction:

Combined gas law tells that the Kelvin temperature of gas is directly proportional to the pressure and volume of a fixed amount of gas.  The mathematical expression for combined gas law can be represented as follows,

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Explanation of Solution

Record the given data,

$\begin{array}{l}{P}_1=735mmHg\\ V_1=\text{7}\text{.31}L\\ T_1=45°Cconvertedto318K\\ P_2=325mmHg\\ T_2=45°Cconvertedto318K\\ V_2=?\end{array}$

Now, substitute these values in rearranged combined gas law and do some simple mathematical calculation to gent final answer as follows,

$V_2=V_1\times \frac{P_1}{P_2}\times \frac{T_2}{T_1}$

$V_2=7.31L\times \left(\frac{735mmHg}{325mmHg}\right)\times \left(\frac{318K}{318K}\right)=16.5L$

The final volume in liters for a sample of carbon monoxide gas at given set of conditions is $16.5L.$

(d)

Interpretation Introduction

Interpretation:

The pressure in atmosphere for a sample of carbon monoxide gas at given set of conditions by using combined gas law has to be determined.

Concept Introduction:

Combined gas law tells that the Kelvin temperature of gas is directly proportional to the pressure and volume of a fixed amount of gas.  The mathematical expression for combined gas law can be represented as follows,

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Explanation of Solution

Record the given data,

$\begin{array}{l}{P}_1=735mmHgconvertedto0.96atm\\ V_1=\text{7}\text{.31}L\\ T_1=45°Cconvertedto318K\\ P_2=?\\ T_2=325°Cconvertedto598K\\ V_2=2.31L\end{array}$

Now, substitute these values in rearranged combined gas law and do some simple mathematical calculation to gent final answer as follows,

$P_2=P_1\times \frac{V_1}{V_2}\times \frac{T_2}{T_1}$

$P_2=0.96atm\times \left(\frac{7.31L}{2.31L}\right)\times \left(\frac{598K}{318K}\right)=5.76atm$

The final pressure in atmosphere for a sample of carbon monoxide gas at given set of conditions is $5.76atm.$