Chapter 10, Problem 40E

Interpretation Introduction

Interpretation:

When sample of ammonia gas has a volume of $\text{50}\text{.0}\text{mL}$ at STP, the Celsius temperature if the volume is $\text{350}\text{.0}\text{mL}$ at $\text{350}\text{mm}\text{Hg}$, is to be calculated.

Concept introduction:

The combination of Charles’s law, Boyle’s law and Gay-Lussac’s law is known as combined gas law. The law relates one thermodynamic variable with the other, and other variables are kept constant.

The combined gas law can be mathematically expressed as shown below.

$\frac{\text{PV}}{\text{T}}=\text{K}$

Answer to Problem 40E

When the volume is $\text{350}\text{.0}\text{mL}$ at $\text{350}\text{mm}\text{Hg}$, the calculated Celsius temperature is $607.06\text{°C}$.

Explanation of Solution

The combined relation between the initial and final pressure, volume and temperature of gas is shown below.

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

Where,

• ${\text{P}}_{\text{1}}$ is the initial pressure.

• ${\text{V}}_{\text{1}}$ is the initial volume.

• ${\text{T}}_{\text{1}}$ is the initial temperature.

• ${\text{P}}_{\text{2}}$ is the final pressure.

• ${\text{V}}_{\text{2}}$ is the final volume.

• ${\text{T}}_{\text{2}}$is the final temperature.

Rearrange the above equation for the value of the final temperature ${\text{T}}_{\text{2}}$.

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

At STP, the temperature is $0\text{°C}$ and the pressure is $\text{1}\text{atm}$.

Conversion of temperature from Celsius to Kelvin can be done as shown below.

$\text{T}\left(\text{K}\right)=\text{T}\left({}^{\text{o}}\text{C}\right)+273$

Therefore, $\text{0}\text{°C}$ is converted to Kelvin as shown below.

$\begin{array}{rll}{\text{T}}_{\text{final}}\left(\text{K}\right)& =& \left(\text{0}+273\right)\text{K}\\ & =& \text{273}\text{K}\end{array}$

Conversion of volume from $\text{atm}$ into $\text{mm}\text{Hg}$ is shown below.

$\text{1 atm}=\text{760 mm}\text{Hg}$

The value of ${\text{P}}_{\text{1}}$ is $\text{760}\text{mm}\text{Hg}$.

The value of ${\text{V}}_{\text{1}}$ is $\text{50}\text{mL}$.

The value of ${\text{P}}_{\text{2}}$ is $\text{350}\text{mm}\text{Hg}$.

The value of ${\text{V}}_{\text{2}}$ is $\text{350}\text{mL}$.

Substitute the values of ${\text{P}}_{\text{1}},{\text{V}}_{\text{1}},{\text{P}}_{\text{2}},{\text{V}}_{\text{2}}$ and ${\text{T}}_{\text{1}}$ in the equation (2).

$\begin{array}{rll}{\text{T}}_{\text{2}}& =& \frac{\text{350}\text{mm}\text{Hg}\times \text{350}\text{mL}\times \text{273}\text{K}}{\text{760}\text{mm}\text{Hg}\times \text{50}\text{mL}}\\ & =& 880.06\text{K}\end{array}$

Conversion of temperature from Kelvin to Celsius can be done as shown below.

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

Therefore, $880.06\text{K}$ is converted to Celsius as shown below.

$\begin{array}{rll}\text{T}\left(\text{°C}\right)& =& \left(880.06-273\right)\text{°C}\\ & =& 607.06\text{°C}\end{array}$

Therefore, the final temperature is $607.06\text{°C}$.

Conclusion

The calculated Celsius temperature, when the volume is $\text{350}\text{.0}\text{mL}$ at $\text{350}\text{mm}\text{Hg}$ is $607.06\text{°C}$.