The currently accepted R value is 8.314J/mol K. Use this number to determine the

Question

The currently accepted R value is 8.314J/mol K. Use this number to determine the molar volume of an ideal gas at room temperature (298.15K) and pressure (1 bar). Present your result with appropriate unit for molar volume. Acid mine drainage, a iron sulfate-sulfuric acid solution system produced by the oxidation of iron sulfate minerals is one of favorite geochemical system to study because it is so easy to observe the geochemical reactions as they occur in the field. For example, the oxidation of pyrite generates a lot of heat so the geochemical gradient in waste pile where pyrite oxidation is occurring is often quite high (Figure 1). It is often possible to identify pyrite-rich areas in mine waste using measurement Temperature and heat production profiles for a pyritic waste pile where the pyrite is undergoing . From Daniel J. A., Marries J. R; and Ritchie A. I. M. (1980). Temperature distributions in an burden dump undergoing pyritic oxidation. Bigeochcmistry of Ancient and Modem Environments. Estimate the geothermal gradient (degree C/Km) for the top 3 m of the waste pile describe in Figure 1 Use thermodynamic data provided at 298.15 K degree and 1 bar(10^5 pascals) pressure, calculate the amount of heat released by the oxidation of one mole of pyrite by the following reaction FeS_2 +7/2 O_2 + H_2O Fe^2+ + 2SO_4^2- + 2H^+ This reaction generates a lot of heat. Some of the heat is conducted out to the surface of the pile but I think a lot of it is used to evaporate liquid water, causing the pile deposition of dissolved iron sulfate minerals. Calculate how many grams of liquid water could be evaporated by the heat from oxidation of 1 g of pyrite. Obviously the oxidation of pyrite is a spontaneous process. Use the thermodynamic data provided to calculate the standard free energy of this reaction. What is the entropy of reaction for pyrite oxidation

Explanation / Answer

I will answer question 1, cause I don't have very much knowledge in the subject of question 2.

first convert pressure from 1 bar to atm:

1 bar * 0.9865 atm/bar = 0.9865 atm

Now let's convert this value into Joules (cause R is in Joules):

0.9865 atm * 101.32 J/L = 99.95 J/L

next, write the expression for an ideal gas with molar volume:

PV = nRT

PV/n = RT

P (V/n) = RT

P Vm = RT

RT/P = Vm

Vm = 8.314 J/mol K * 298.15 K / 99.95 J/L All the units of K and J goes away.

Vm = 24.8 L/mol

Usually the molar volume is m3 so:

Vm = 24.8 L/mol * 1 m3/1000 L = 0.0248 m3/mol

I apologize for not provide an answer for question 2, but at least you have this.

Hope this helps

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