CHM 1045 CH5 Homework Review example 5.4 and exercise 5.4. Suppose 10.0 L of a g

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CHM 1045 CH5 Homework Review example 5.4 and exercise 5.4. Suppose 10.0 L of a gas at 1.10 atm and 25.0 °C is heated to 100.0°C and allowed to expand until a final pressure of 1.20 atrm is reached. Rearrange the combined gas law and use it to determine the final volume. Show all units. (2 pts) 1. Convert the final pressure for problem #1 into mmHg, Pa (kg/m-s2), and kPa Also, use 1 atm = 14.696 psi to convert to psi (pounds per square inch). Refer to conversion factors in the chapter 5 notes. Show all units and conversion factors. (2 pts) 2. Review example 5.7 and exercise 5.7. Suppose there are 2.00 moles of O2 at 1520 mm Hg and 27.0°C, Rearrange the ideal gas law and use it to determine the volume (L) of the gas. Then, find the density by rearranging the equation P(Mm)- dRT. Show all units. (2 pts) 3.

Explanation / Answer

First condition, V1= 10L, P1= 1.1 atm, T1= 25 deg.c= 25+273= 298K. Assuming heating to 100 deg.c (T2= 100+273=373K) is done at constant pressure, the gas law equation, P1V1/T1= P2V2/T2 becomes

V1/T1= V2/T2, V2= V1T2/T1= 10*373/298 L=12.52 L

Now while keeping the temperature constant, pressure from P2= 1.1 atm is raised to P3= 1.2 atm, V2= 12.52L, from gas law, P2V2= P3V3, V3= P2V2/P3= 1.1*12.52/1.2 =11.47 L

2. final pressure = 1.2 atm, 1 atm= 760 mm Hg, 1.2 atm= 1.2*760 mm Hg= 912 mm Hg

1 atm= 101.3 Kpa, 1.2 atm = 101.3*1.2 Kpa =121.56 Kpa, 1 Kpa= 1000Pa, 121.56 Kpa =121.56*1000 Pa

1 atm= 14.696 inches of Hg

1.2 atm= 1.2*14.696 inches of Hg =17.6352 inches of Hg.

3. From gas law, PV= nRT, given n= no of moels of oxygen =2 moles, P= pressure in atm , 1 atm= 760 mm Hg

1520 mm Hg= 1520/760 atm =2 atm, T= 27 deg.c= 27+273= 300K, R =0.0821 L.atm/mole.K

V= nRT/P= 2*0.0821*300/2=24.63 L

density = mass/Volume, molar mass of O2= 2 moles, mass of O2= moles* molar mass= 2*32= 64 gm

density = 64/24.63 L= 2.598 g/L

4. Vanderwaal constants for O2, a= 1.382 L2 bar/mole2, b= 0.03186 L/mole, n=1, V=22.5 L, T= 27deg.c= 27+273= 300K, R= 0.08206 L.bar/mole.K

from gas law for one moles, P= RT/(V-b)- a/V2 = 0.08206*300/(22.5-0.03186)- 1.382/(22.5*22.5)=1.095-0.00273= 1.093 bar

5. Mole fraction =moles/total moles

mole fractrions: N2= 1.58/(1.58+0.42)= 1.58/2=0.79 and O2= 0.42/(1.58+0.42)= 0.21

partial pressure= mole fraction* total pressures

partial pressures : N2= 0.79*3=2.37 atm, O2= 0.21*3= 0.63 atm

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