PHY 3513: Thermodynamics Midterm Examination Solutions, Exams of Thermal Physics

Material Type: Exam; Class: THERMAL PHYSICS; Subject: PHYSICS; University: University of Florida; Term: Unknown 1989;

Typology: Exams

Pre 2010

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February 11, 2005 Name: -----------------------------.
PHY 3513:Kumar
I Midterm Examination: Solutions
1) A tank of volume 1.0 m3 contains oxygen at a pressure of 1.5×106 Pa and a
temperature of 20° C. Assume that oxygen behaves like an ideal diatomic gas.
a) How many kilomoles of oxygen are in the tank?
n=P1V1 /RT = 1.5
×
106
×
1/(8.314
×
103
×
293) = 0.616 kilomoles
b) How many kilograms?
m=n
×
32kg/kilomoles = 19.7kg
c) Find the pressure if the temperature is raised to 500° C.
P2=(P1/T1)
×
T2 = (773/293)
×
1.5
×
106 = 3.96
×
106Pa
d) At 20°C, how many kilomoles can be withdrawn form the tank before the
pressure falls to 10% of the original pressure.
Δ
n = 0.9PV/RT = 0.9
×
1.5
×
106
×
1/8.314
×
103
×
293 = 0.554 kilomoles
2) Show that for a van der Waals’ gas β/κ = R/(v-b).
Here please note that
β
/
κ
= dP/dT|v = R/(v-b).
3) In the cycle shown, P1 = 10 atm, V1 = 2 m3 , V2 = 4 m3 and n= 1 kilomole. The gas is
diatomic and the step 1Æ 2 is adiabatic.
i) What is T2?
ii) How much work does this engine do?
iii) What is the efficiency of this engine?
The parameters of the system are: T1 = P1 V1 /nR =
10
×
1.013
×
105
×
2/8.314
×
103=243.69K
Also P2 = P1(V1/V2)
γ
=10.13
×
105/21.4=3.84
×
105Pa; T3 = P2V1/nR = 92.34K.
1. Since 1
Æ
2 is adiabatic and the gas is diatomic (
γ
= 1.4), T2=(V1/V2)0.4 T1 =
184.68K
2. Qh=ncv(T1-T3) = (5/2)8.314
×
103(243.69-92.34) = 3.15
×
106 J
|Qc|=ncp(T2-T3) = (7/2)8.314
×
103(184.68-92.34) = 2.69
×
106 J
W = Qh-Qc = 0.46
×
106 J.
3.
η
= W/Qh = 0.146
14.6%.
1
2
P
V
pf2

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February 11, 2005 Name: -----------------------------.

PHY 3513:Kumar

I Midterm Examination: Solutions

  1. A tank of volume 1.0 m^3 contains oxygen at a pressure of 1.5× 106 Pa and a temperature of 20° C. Assume that oxygen behaves like an ideal diatomic gas.

a) How many kilomoles of oxygen are in the tank?

n=P 1 V 1 /RT = 1.5 × 106 ×1/(8.314 × 103 ×293) = 0.616 kilomoles

b) How many kilograms?

m=n ×32kg/kilomoles = 19.7kg

c) Find the pressure if the temperature is raised to 500° C.

P 2 =(P 1 /T 1 ) ×T 2 = (773/293) ×1.5× 106 = 3.96 × 106 Pa

d) At 20°C, how many kilomoles can be withdrawn form the tank before the pressure falls to 10% of the original pressure.

Δn = 0.9PV/RT = 0.9 ×1.5× 106 ×1/8.314 × 103 ×293 = 0.554 kilomoles

  1. Show that for a van der Waals’ gas β/κ = R/(v-b).

Here please note that β/ κ = dP/dT| v = R/(v-b).

  1. In the cycle shown, P 1 = 10 atm, V 1 = 2 m^3 , V 2 = 4 m^3 and n= 1 kilomole. The gas is diatomic and the step 1Æ 2 is adiabatic.

i) What is T 2? ii) How much work does this engine do? iii) What is the efficiency of this engine?

The parameters of the system are: T 1 = P 1 V 1 /nR =

10 ×1.013× 105 ×2/8.314 × 103 =243.69K

Also P 2 = P 1 (V 1 /V 2 )^ γ^ =10.13 × 105 /21.4=3.84 × 105 Pa; T 3 = P 2 V 1 /nR = 92.34K.

1. Since 1 Æ2 is adiabatic and the gas is diatomic ( γ = 1.4), T 2 =(V 1 /V 2 )0.4^ T 1 =

184.68K

2. Qh =nc v(T 1 -T 3 ) = (5/2)8.314 × 103 (243.69-92.34) = 3.15 × 106 J

|Qc|=nc p (T 2 -T 3 ) = (7/2)8.314 × 103 (184.68-92.34) = 2.69 × 106 J

W = Qh -Qc = 0.46 × 106 J.

3. η = W/Qh = 0.146 ⇒ 14.6%.

P

V

  1. Forty kg of water at 0°C must be frozen into ice in a refrigerator. The room temperature is 25° C. The latent heat of fusion of water is 3.33× 105 J/kg. What is the minimum power required if the freezing has to occur in an hour?

Minimum power must come from the most efficient refrigerator, which is a Carnot cycle. The coefficient of performance of a Carnot refrigerator is c = Qc/W = Tc/(Th - Tc ).

Therefore the power 3600

= × ×

= (^) c c

h cQ T

T T

P =339W.