   Chapter 10, Problem 159IP

Chapter
Section
Textbook Problem

One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime (CaO) at 600.°C followed by condensation of the mercury vapor. Given the heat of vaporization of mercury (296 J/g) and the vapor pressure of mercury at 25.0°C (2.56 × 10−3 torr), what is the vapor pressure of the condensed mercury at 300.°c? How many atoms of mercury are present in the mercury vapor at 300.°c if the reaction is conducted in a closed 15.0-L container?

Interpretation Introduction

Interpretation

At 300°C the vapor pressure of condensed Mercury and number of Mercury atoms in Mercury vapor to be calculated.

Concept Introduction

Vapor pressure: In a closed system, at given temperature the pressure evolved by a vapor in thermodynamic equilibrium with its solid or liquid phases (condensed phase).

ln(P1P2)ΔHVapR[1T2-1T1]

Where,

P1 and P2 arepressure;T1 and T2 are Temperature.

Ideal gas law: This law expressed by following equation is,

PV = nRT

Where,

P - PressureV - volumen - Number of molesR - GasconstantT -Temperature

Explanation

Explanation

Record the given data

Temperature involved in preparing elemental Mercury = 600°C

Heat of vaporisation of Mercury = 296J/g

Vapour pressure of Mercury = 2.56×10-3 at 25°C

Temperature involved in preparing elemental Mercury, heat of vaporization and vapor pressure of Mercury are recorded as shown.

To calculate molality of Mercury.

ln(P1P2)ΔHVapR[1T2-1T1]

= 296Jg× 200.6gmol= 5.94×104J/mol Mercury

By plugging the value of heat of vaporization of Mercury and temperature, a mole of Mercury has calculated.

calculate vapor pressure of condensed Mercury.

ln[2.56×10-3 torrP2]=5.94×104J/mol8

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