Chapter 24, Problem 100A

Interpretation Introduction

Interpretation:

Volume of gas at STP when volume is $\text{4}\text{.5}\text{L}$ at pressure $\text{0}\text{.65}\text{atm}$ and temperature $\text{321}\text{K}$ is to be explained.

Concept introduction:

Ideal gases are composed of large number of particles moving randomly are assumed to have perfect elastic collisions among themselves. It is a theoretical concept. Gases that show perfect elastic collision are practically not possible. At higher $\text{T}$ and lower $\text{P}$ gases behave almost ideally.

Ideal gas law is an equation of state for any hypothetical gas. Expression for ideal gas equation is as follows:

$\text{PV}=\text{nRT}$

There,

$\text{P}$ is pressure of the gas.

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

$\text{n}$ denotes moles of gas.

$\text{R}$ is gas constant.

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

Answer to Problem 100A

Volume of gas at STP when volume is $\text{4}\text{.5}\text{L}$ at pressure $\text{0}\text{.65}\text{atm}$ and temperature $\text{321}\text{K}$ is $\text{2}\text{.488}\text{L}$.

Explanation of Solution

Ideal gas equation is also expressed as follows:

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}{\text{T}}_{\text{2}}}$

Here,

${\text{P}}_{\text{1}}$ isinitial pressure of the gas.

${\text{P}}_{\text{2}}$ isfinal pressure of the gas.

${\text{V}}_{\text{1}}$ isinitial volume of gas.

${\text{V}}_{\text{2}}$ isfinal volume of gas.

${\text{n}}_{\text{1}}$ denotes the initial number of moles of gas.

${\text{n}}_{\text{2}}$ denotes the initial number of moles of gas.

${\text{T}}_{\text{1}}$ isinitial temperature of gas.

${\text{T}}_{\text{2}}$ isfinal temperature of gas.

Here number of moles remain constant so ${\text{n}}_{\text{1}}$ is equal to ${\text{n}}_{\text{1}}$.

So the modified ideal gas equation is as follows:

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}=\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

Rearrange above equation to calculate ${\text{V}}_{\text{2}}$.

${\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}{\text{P}}_{\text{2}}}$

At STP, ${\text{T}}_2$ is $\text{273}\text{.15}\text{K}$ and ${\text{P}}_2$ is $\text{1}\text{atm}$.

Substitute $\text{4}\text{.5}\text{L}$ for ${\text{V}}_1$, $\text{0}\text{.65}\text{atm}$ for ${\text{P}}_1$, $\text{321}\text{K}$ for ${\text{T}}_1$, $\text{273}\text{.15}\text{K}$ for ${\text{T}}_2$ and $\text{1}\text{atm}$ for ${\text{P}}_2$ in above equation to calculate ${\text{V}}_{\text{2}}$.

$\begin{array}{c}{\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}{\text{P}}_{\text{2}}}\\ =\frac{\left(\text{0}\text{.65}\text{atm}\right)\left(\text{4}\text{.5}\text{L}\right)\left(273.15\text{K}\right)}{\left(\text{321}\text{K}\right)\left(\text{1}\text{atm}\right)}\\ =\text{2}\text{.488}\text{L}\end{array}$

Volume of gas at STP when volume is $\text{4}\text{.5}\text{L}$ at pressure $\text{0}\text{.65}\text{atm}$ and temperature $\text{321}\text{K}$ is $\text{2}\text{.488}\text{L}$.

Conclusion

Volume of gas at STP when volume is $\text{4}\text{.5}\text{L}$ at pressure $\text{0}\text{.65}\text{atm}$ and temperature $\text{321}\text{K}$ is $\text{2}\text{.488}\text{L}$.