Chapter SM, Problem 14PQ

Interpretation Introduction

Interpretation:

Volume of dry gas at STP has to be determined.

Concept Introduction:

Ideal gas is hypothetical or theoretical gas that is formed from large number of randomly moving particles with perfectly elastic collisions between them. This gas is considered to follow ideal gas law under all conditions. It is used to predict behavior and properties of various gases under certain conditions. It represents relationship between pressure, temperature, volume and amount of gas.

The formula for ideal gas equation is as follows:

  $\text{PV}=\text{RT}$        (1)

Here,

  $\text{P}$ is pressure of the gas.

  $\text{V}$ is volume of gas.

  $\text{T}$ is temperature of the gas.

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

Answer to Problem 14PQ

Option (B) is the correct one.

Explanation of Solution

Reason for correct option:

For gas at temperature ${\text{T}}_1$, pressure ${\text{P}}_1$ and volume ${\text{V}}_1$; equation (1) modifies as follows:

  ${\text{P}}_1{\text{V}}_1={\text{RT}}_1$        (2)

Here,

  ${\text{P}}_1$ is pressure of gas over water.

  ${\text{V}}_1$ is volume of gas over water..

  ${\text{T}}_1$ is temperature of gas over water.

For gas at temperature ${\text{T}}_2$, pressure ${\text{P}}_2$ and volume ${\text{V}}_2$; equation (1) modifies as follows:

  ${\text{P}}_2{\text{V}}_2={\text{RT}}_2$        (3)

Here,

${\text{P}}_2$ is pressure of dry gas.

${\text{V}}_2$ is volume of dry gas.

${\text{T}}_2$ is temperature of dry gas.

Divide equation (1) by equation (2) as follows:

  $\frac{{\text{P}}_1{\text{V}}_1}{{\text{P}}_2{\text{V}}_2}=\frac{{\text{T}}_1}{{\text{T}}_2}$        (4)

Rearrange equation (4) to calculate ${\text{V}}_2$.

  ${\text{V}}_2=\frac{{\text{P}}_1{\text{V}}_1{\text{T}}_2}{{\text{P}}_2{\text{T}}_1}$        (5)

Temperatures are to be converted into $\text{K}$. Conversion factor for this is as follows:

  $0°\text{C}=273\text{ K}$

Therefore, value of ${\text{T}}_2$ is $273\text{ K}$.

Therefore, ${\text{T}}_1$ can be calculated as follows:

  $\begin{array}{c}{\text{T}}_1=\left(25+273\right)\text{ K}\\ =\text{298 K}\end{array}$

Conversion of pressure from $\text{atm}$ to $\text{mm Hg}$ is as follows:

  $1\text{ atm}=760\text{ mm Hg}$

Therefore, value of ${\text{P}}_2$ is $760\text{ mm Hg}$.

The pressure of dry gas $\left({\text{P}}_1\right)$ can be calculated as follows:

  $\begin{array}{c}{\text{P}}_1=\left(\text{Pressure of gas collected over water}-\text{Pressure of water}\right)\\ =\left(740-24\right)\text{ mm Hg}\end{array}$

Substitute $\left(740-24\right)\text{ mm Hg}$ for ${\text{P}}_1$, $500\text{ mL}$ for ${\text{V}}_1$, $298\text{ K}$ for ${\text{T}}_1$ and $760\text{ mm Hg}$ for ${\text{P}}_2$ and $273\text{ K}$ for ${\text{T}}_2$ in equation (5).

  ${\text{V}}_2=\frac{\left(740-24\right)\left(500\right)\left(273\right)}{\left(760\right)\left(298\right)}$

Hence option (B) is correct.

Reason for incorrect option:

Volume of dry gas becomes $500\times \frac{740-24}{760}\times \frac{273}{298}$ that does not match with $500\times \frac{740+24}{760}\times \frac{298}{273}$, $500\times \frac{760}{740-24}\times \frac{273}{298}$ and $500\times \frac{760}{740+24}\times \frac{298}{273}$. Hence option (A), (C) and (D) are incorrect.

Conclusion

Volume of dry gas at STP is $500\times \frac{740-24}{760}\times \frac{273}{298}$. Hence, correct option is (B).