Exercises in OpenStax Chemistry, Ch. 9

Ryan Kilby

CHM/150

April 1, 2017

Dr. ELIZABETH FRAYNE

 

 

 

Week Two Exercises in OpenStax Chemistry, Ch. 9

           

            Chapter Nine

6. A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?

 

So we know that barometric pressure is P = 740 torr

So that means that 1 torr is 0.00132 atm

Pressure is 740 x 0.00132

                = 0.9768 atm

Pressure is 740 mmHg

Pressure is 740 * 0.1333

               is 98.64 kPa

 

 

 28. What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr?

 

So in this case V1/T1=V2/T2

So that means that 13.3-L/328-K= 11.2-L/T2

This gives us T2= 11.2*328-k/13.2-k

With an end temperature of T2=276.21-K

 

36. A high altitude balloon is filled with 1.41 × 104 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?

 

In this case P1V1/T1 is equal to P2V2/T2

V2 is P1V1T1/T1P2 = 745torr*1.41 × 10^4 L*225K/ (63.1torr*294K) = 127.4L

 

 

48. What is the density of laughing gas, dinitrogen monoxide, N2O, at a temperature of 325 K and a pressure of 113.0 kPa?

 

P is equal to 113KPa

T is 325k

M = 44

 Gas constant is R = 8.314J/ Mol K

pv = nrt

n = m/M

so the density = m/v 

p = density* RT/M

so density is =  PM/RT

          (44*113)/ (8.314*325)

          Is equal to1840.08Kg/m^3 or 1.84g/l

 

 

58. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2, and the remainder N2 at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)

 

 

 

 

84. Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H2O and D2O.

 

1 / x = square root of (20.03 / 18.01)

1 / x = 1.054

x = 1 / 1.054 = 0.95

 

94. What is the ratio of the average kinetic energy of a SO2 molecule to that of an O2 molecule in a mixture of two gases? What is the ratio of the root mean square speeds, urms, of the two gases?

 

kinetic energy is 3kT/so the ratio will be 1:1

urms1/urms2 = sqrt(M2/M1)

urmsSO2/urmsO2 = sqrt(16/64) = aqrt(1/4) = 1/2 = 0.5

 

104. A 0.245-L flask contains 0.467 mol CO2 at 159 °C. Calculate the pressure:

(a) Using the ideal gas law

(b) Using the van der Waals equation

(c) Explain the reason for the difference.

(d) Identify which correction (that for P or V) is dominant and why.

 

n = 0.467moles

v = 0.245L

t = 159+273 = 432K

Gas constant is R = 8.314J/Mol.K

1. Pv = nrt

P = 0.467* 8.314*432/0.245

P =6846.12 Pascals

a = 3.64, b = 0.04267

From above, P = 9 (Nrt) / (v-nb)- n^2 a/v^2

P = 7439.02pascals

3. The difference is carried by correction of intermolecular forces so constant a0 and correction of finite molecular size constant b.
4. The variation of the pressure is dominant, so the volume of the flask is fixed. But this will be subject to change in the event another flask is used.

 

 

 

 

 

 

 

 

 

 

 

 

 

Reference

Rice University. (2015). OpenStax College. Retrieved from Rice University, Chemistry website.

 

 

 

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