Chemistry 114
First Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1. According to Wikipedia at the bottom of the Mariana trench (~6.831 miles below sea
level) the pressure is 1086 bar.
A. Convert this pressure to ATM
B. Convert this pressure to PSI
C. Convert this pressure mm Hg
(mm Hg = torr!)
2. I am going to fill a 1.5 L balloon with air here in Spearfish where the pressure is 665 torr
and the temperature is 12 oC, and then take it to the top of Harney Peak where the
pressure is 581 mm Hg and the temperature is 5oC. What is the volume of the balloon
now?
PV=nRT; P1V1/T1 =nR= P2V2/T2
Or you can solve for n under condition 1 and plug in for condition 2.
I’ll go with the first equation so I don’t have to convert P into atm
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3. What is the density of neon gas when the pressure is 15 atm and the temperature is
200oC?
Density=(P/RT)×Molar mass
Density of gases is usually expressed in g/liter
Density of liquids and solids in g/mL
4. I have a 3L container of H2 gas at 3 Atm of pressure. I will connect this with a 4L
container of O2 gas at 4 Atm of pressure. I will then open a valve between the two
containers and let the gases mix.
A. What is the partial pressure of O2 in the final gas mixture?
Expand 4L of O2 @4 atm into a 7L total volume (P 1V 1=P 2V 2 ; P 2 = P 1(V1/V2)
4 atm ×(4L/7L) = 2.29 atm
B. What is the total pressure of the gases in the final mixture?
Do the same for H2: 3 atm ×(3L/7L) = 1.29 atm
Sum the two pressures P=2.29 + 1.29 = 3.58 atm
C. What is the mole fraction of H2 in the final gas mixture?
×=P1/PTot = 1.29/3.58 = .36 (no units)
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5.Define (or give an equation for) the following terms:
An Intensive property
A property that does not depend on the amount of material in the system.
A path or transfer function
A function that depends on the path taken between two states.
H
Enthalpy = U+PV
qV
Heat evolved under constant volume conditions =ÄE
Molar heat capacity
The amount of heat needed to raise the temperature of 1 gram of material by 1 oC (or K)
The heat gained or lost when a compound in its standard state is formed from its
elements in their standard states.
Adiabatic
No heat is exchanged between the system and its surroundings.
6. I am going to transfer 2.5 kJ of energy into a system. At the same time the system will
expand from 3L to 15 L against a pressure of .3 atm. What is the ÄE of the system.
ÄE=q+w; 2.5 kJ=q; w= -PÄV
ÄV = 15-3 = 12R
-PÄV= -.25×12R = -3.0RAatm
-3.0RAatm×(101.315J/RAatm) = -360J
ÄE= 2500J -360J = 2140J
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7. Using these reactions:
H2(g) + ½ O2(g) 6 H2O(l)
N2O5(g) + H2O(l)6 2 HNO3(l)
½ N2(g) + 3/2 O2(g) + ½ H2(g) 6 HNO3(l)
Determine ÄHrxn for the reaction:
ÄH1 -295.8
ÄH2 -66.6
ÄH3 -184.1
2N2(g) + 5O2 (g)6 2N2O5(g)
To get 2 N2O5 as a product, reverse RXN 2 and multiply by 2
4 HNO3(l) 62N2O5(g) + 2H2O(l)
-2× -66.7
+133.2
To get rid of the 4HNO3 reactants, multiply RXN 3 by 4
2 N2(g) + 6 O2(g) + 2 H2(g) 6 4 HNO3(l)
4× -184.1
-736.4
To get rid of the 2 H2O product, reverse RXN 1 and multiply by 2
2 H2O(l)62 H2(g) + 1 O2(g)
-2× -295.8 +591.6
Sum it all up
4 HNO3(l) +2 N2(g) + 6 O2(g) + 2 H2(g) + 2 H2O(l)
6 2 H2(g) + 1 O2(g) + 4 HNO3(l) + 2N2O5(g) + 2H2O(l)
2N2(g) + 5O2 (g)6 2N2O5(g)
+133.2-736.4+591.6= -11.6
8. I have a 100g block of gold that is at 90oC, and I place it in contact with a 100g block of
a second metal that is at 60oC in a well insulated container so the heat is exchanged
between the two blocks under adiabatic conditions. If the final equilibrium temperature of
the two blocks is 75oC, and the specific heat capacity of gold is .129J/oCAg, what is the
specific heat capacity of the second metal?
Adiabatic so heat lost by warm + heat gained by cold =0
Heat lost by warm = 100 g ×.129J/oCAg ×(75-90)
Heat gained by cold = 100g × X J/oCAg ×(75-50)
0=[100 g ×.129J/oCAg ×(-15oC)] +[ 100g × X ×(15oC)]
= -193.5J + 1500X gAoC
193.5J/1500 gAoC =X
X= .129 J/gAoC
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