Chapter 10, Problem 76GQ

Interpretation Introduction

Interpretation: For the two given set of gases each with given volume and pressure the partial pressure of each gas and the total pressure of the gas after valve has to be opened.

Concept introduction:

Ideal gas Equation:

Any gas can be described by using four terms namely pressure, volume, temperature and the amount of gas.  Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained.  It is referred as ideal gas equation.

  $\begin{array}{l}\text{V }\propto \frac{\text{nT}}{\text{P}}\\ \text{V = R}\frac{\text{nT}}{\text{P}}\\ \text{PV = nRT}\\ \text{where,}\\ \text{n = moles of gas}\\ \text{P = pressure}\\ \text{T = temperature}\\ \text{R = gas }\text{constant}\end{array}$

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties.  At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to the volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Avogadro’s Hypothesis:

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

Van der Waal’s gas equation:

The van der Waal equation describes the ideal gas as it approaches to zero.  The van der Waal equation contains correction terms a and b for the intermolecular forces and molecular size respectively.

The van der Waal equation is as follows,

  $\left[P+a{\left(\frac{n}{V}\right)}^2\right]\left(\frac{V}{n}-b\right)=RT$

Answer to Problem 76GQ

The partial pressure of $\text{He = 87}\text{.0mm Hg}$ and the partial pressure of $\text{Ar = 142 mm Hg}$.  The total pressure of the gas after the valve opened is equal to $\text{229 mm Hg}$

Explanation of Solution

Given,

  $\begin{array}{l}\text{Before mixing}\\ \text{He}\text{Ar}\\ \text{V}\text{ 3}\text{.0 L}\text{2}\text{.0 L}\\ \text{P}\text{145 mm Hg}\text{335 mm Hg}\end{array}$

The partial pressure for the given gases is determined as follows,

  $\begin{array}{l}\text{Partial pressure for He =}\frac{\text{ initial pressure for He}\times \text{Initial volume of He}}{Totalvolume}\\ =\frac{\text{ 145mm Hg}\times \text{3L}}{5L}=87mmHg\end{array}$

  $\begin{array}{l}\text{Partial pressure for Ar =}\frac{\text{ initial pressure for Ar}\times \text{Initial volume of Ar}}{Totalvolume}\\ =\frac{\text{ 355mm Hg}\times 2\text{L}}{5L}=142mmHg\end{array}$

The total pressure for a gas is determined by sum of all the partial pressure of the gas involved in a chemical reaction.

  $\begin{array}{l}{\text{Total pressure = He}}_{\text{partial pressure}}+A{r}_{partialpressure}\\ =87mmHg+142mmHg=229mmHg\end{array}$

Conclusion

The partial pressure of each gas present in the given chemical reaction and the total pressure of the gases were determined.