Thermodynamics Problems and Solutions: Ideal Gas Law
Ideal Gas Law and Thermodynamics Problems
Problem 1
An ideal gas occupies a volume of 100 cm3 at 20°C and a pressure of 100 Pa. Determine the number of moles of gas in the vessel.
1 Pa * 1 m3 = 1 Joule
PV = nRT
100 Pa * 10 * 10-7m3 = n
8.31 * 293
n = 4.11 * 10-6
Problem 2
Calculate the volume occupied by one mole of gas under normal conditions.
PV = nRT
1 atm * V = 1 mol * 0.082 * 273
V = 22.4 L
Problem 3
Helium is introduced into a mobile container. The initial pressure, volume, and temperature of the gas are 15 L, 200 kPa, and 27°C. If the volume is reduced to 12 L and the pressure increases to 350 kPa, find the final temperature (Tf) of the gas.
P1V1 = P2V2
T1 T2
1 atm – 101,325 Pa
1.97 atm * 15 = 3.45 * 12
300 K T2
T2 = 420 K
Problem 4
The rim of a bicycle is filled with air at a gauge pressure of 550 kPa and 20°C. What is the gauge pressure of the tire after a ride on a hot day when the air temperature of the rim is 40°C?
P1/T1 = P2/T2
5.42 / 293 = P2 / 313
P2 = 5.78 atm
P2 = 586 kPa
Problem 5
7.5 liters of oxygen is compressed adiabatically until its volume is reduced to 1 L. At the end of compression, the pressure established is 1.6 * 106 Pa. What was the pressure before the gas compression?
P1V1 = P2V2
P1 = P2V2 / V1
P1 = 1.6 * 106 * 11.4 / 7.51.4 = 95287.122 Pa
Problem 6
A Carnot engine whose hot reservoir is 400 K takes 100 cal at this temperature in each cycle and yields 80 calories to the cold reservoir. What is the temperature of the latter?
(Q2 – Q1) / Q2 = (T2 – T1) / T2
100 – 80 / 100 = 400 – T1 / 400
0.2 * 400 = 400 – T1
T1 = 320 K
Problem 7
A Carnot engine operates between two heat sources at temperatures of 400 K and 300 K. If in each cycle, the motor receives 1200 cal from the hot source, how many calories per cycle does it yield to the cold source?
Q1 / Q2 = T1 / T2
Q2 = 900 cal
Problem 8
In a closed container, there are 20 grams of N2 and 32 g of O2. Find the change in internal energy of this gas mixture when cooled by 28°C.
U = n Cv * T
1.71 * 5 * 28 * 4.186
= 1002.1284
Problem 9
WAB = 0, ΔU = Q = 600 J
The change in internal energy
WBD = P(V2 – V1) = 8 * 104 (5 * 10-3 – 2 * 10-3) = 240 J
ΔUABD = QABD – WABD = 800 – 240 = 560 J
Total heat to be supplied if it goes to the initial state from the final state
ΔUACD = 560 J
WAC = 3 * 104 (5 * 10-3 – 2 * 10-3)
WAC = 90 J
QACD = ΔUACD + WACD = 650 J
Problem 10
A gas is in a steel cylinder at 20°C and 5 atm pressure.
A. If the cylinder is surrounded by boiling H2O and allowed to reach equilibrium, what is the heat of the gas?
P1V1 / T1 = P2V2 / T2
P2 = 6.36 atm
B. If gas escapes until the pressure is again 5 atm, what weight fraction of the original gas has escaped?
V1 = nRT / P1 = X / 6.36
V2 = nRT / P2
V1 / V2 = 78.6%
Problem 11
A system of gas initially at V = 2 L and P = 3 * 105 dyne/cm2 takes the volume up to P = 8 * 105 dyne/cm2. It is supplied with 600 J of heat. Right away, it takes P = constant up to a V = 5 L. It is supplied with 200 J of heat.
a. ΔV in the first process
Wab = 0, ΔV = Q = 600 J
b. ΔV in the whole transformation
Wbd = P(V2 – V1) = 8 * 104 (5 * 10-3 – 2 * 10-3)
Wbd = 240 J
ΔV = Qabd – Wabd = 800 – 240 = 560 J
c. The total heat supplied to be transferred to go from the initial state to the final one is made, and then P = constant and V = constant
ΔVacd = 560 J
Wac = 3 * 104 (5 * 10-3 – 2 * 10-3)
Wac = 90 J
Qacd = ΔVacd + Wacd = 560 + 90
Qacd = 650 J
8 | B —— d
| |
| |
3 | A —– c
————–
2 5
2 = 2 * 10-3
5 = 5 * 10-3
3 = 3 * 104
8 = 8 * 104
Problem 12
To increase the heat radiated twice, it is necessary to increase the body temperature in an absolute:
d) 18.9%
σ = 5.67 * 10-8
H = σ * e * A * T4
2H = ditto above
T1 = 65
T2 = 77.3
T2 = 4√(2H/eA)
65 → 100
77.3 → xx = 18.9%
Problem 13
The relative humidity of air is 37.2%. If it is known that the dew point is 5°C, this is where the air pressure is H2O steam 6.51 mmHg, then the vapor pressure of water at the temperature in which the air is:
b) 17.5 mmHg
Hrelative = partial pressure * 100 / Psteam
37.2 = 6.51 * 100 / Psteam
P = 651 / 37.2 = 17.5 mmHg
Problem 14
Heat is added to a monatomic gas contained in a closed metal container that was initially at 27°C until its pressure increases by 45%. What is the final temperature?
e) 162°C
P1 / P2 = T1 / T2
1P / 1.45P = 300 / T2
T2 = 435 K = 161.8°C
Problem 15
1 kg of air is at a temperature of 20°C and expands at 2.5 atm pressure adiabatically. This happens until its pressure lowers to 1 atm. Then the ratio is V2/V1.
b) 1.92
γ = 1.4
m = 1000 g
T = 293 K
P1 = 2.5 atm
P2 = 1 atm
P1 * V1γ = P2 * V2γ
2.5 * V1γ = 1 * V2γ
V2 / V1 = 1.4√(2.5 / 1) = 1.92
Problem 16
In the previous issue, the final temperature of the system will be:
d) -47°C
T1 * P1(1 – γ / γ) = T2 * P2(1 – γ / γ)
293 K * 2.5(1 – 1.4 / 1.4) = T2 * 1(1 – 1.4 / 1.4)
T2 = 225.5 K = -47°C
Problem 17
In problem 2, the work of gas to expand will be:
e) none
d = m / V
V = 1000 g / (1.2929 * 10-3 g/cm3)
V1 = 773455.02 cm3
V1 = 773.42 L
P1 * V1γ = P2 * V2γ
2.5 * 773.41.4 = 1 * V21.4
27650 = V21.4
V2 = 1.4√(27650) = 1488.15 L
W = (P2V2 – P1V1) / (1 – γ) = (1 * 1488.15 – 2.5 * 773.4) / (1 – 1.4)
W = 112743.8 J
Problem 18
In problem 2, the internal energy change experiment will be:
e) none
W = -112743.8 J
Problem 19
The dew point is the temperature at which:
c) The water vapor in the air becomes saturated steam
Problem 20
A diatomic gas expands at constant pressure, provided 350 cal. Then the work done by the gas is:
a) 42.7 kgm
Problem 21
1 mole of diatomic gas is subject to 3 atm pressure at a temperature of 27°C. After heating at constant pressure, the gas occupies a volume of 30 L.
The amount of heat received by the gas is:
e) any
Q = 23324.3 J
Problem 22
The internal energy change of the gas is approximately:
c) 3980 cal
Problem 23
The work done by the gas is approximately:
1579.4 cal
Problem 24
1 mole of oxygen is initially at 1 atm and a temperature of 0°C. It is first compressed isothermally to a third of its volume and then expands adiabatically to recover its initial pressure.
Work performed in the compression is:
b) -600 cal
Problem 25
Exchange of color in the compression is approximately:
-600 cal
Problem 26
Work performed in the expansion is approximately:
364.9 cal
Problem 27
The variation of internal energy in the expansion is approximately:
ΔU = -364.9 cal
Problem 28
A mole of gas
Pi = 2 atm
Vi = 0.3 L
ΔV = 91 J
In its final state:
Pf = 1.5 atm
Vf = 0.8 L
ΔV = 182 J
Calculate the three paths
2 | i —— B
| |
| |
1.5 | A ——- F
————-
0.3 0.8
WIAF = 1.5 (0.8 – 0.3)
W = PΔV
= 0.75 atm = 76.3 J/L
WIBF = 2 (0.8 – 0.3) = 1 atm = 101.8 J/L
WIF = (1 + 2) / 2 = 0.875 = 89.07 J/L
1 atm = 24.32 cal
Net heat transfer
ΔV = Q – W
QIBF = ΔV + W
(Vf – Vi) + 101.8
91 + 101.8 = 192.8
QIAF = 91 + 76.35 = 197.35
QIF = 91 + 89.07 = 180.07
Problem 29
An air mass of 1000 g is found at a temperature of 30°C and a pressure of 1.5 atm. It expands adiabatically until the pressure drops to 1 atm. Then the temperature at the end of the process will be:
c) -3°C
T1 * P1(1 – γ / γ) = T2 * P2(1 – γ / γ)
303 * 1.5(1 – 1.4 / 1.4) = T2 * 1(1 – 1.4 / 1.4)
T2 = -3°C
Problem 30
In the previous issue, the variation of internal energy experienced in the system will be:
e) none
d = m / V = 1.2929 * 10-3 g/cm3
V = m / d = 1000 / (1.2929 * 10-3) = 773455.02 cm3 = 773.4 L = V1
P1 * V1γ = P2 * V2γ
1.5 * 773.41.4 = V21.4
V2 = 1033.1 L
W = (P2V2 – P1V1) / (1 – γ) = (1 * 1033.1 – 1.5 * 773.4) / (1 – 1.4) = 317.2 atm * L = 32144.2 J = 7677 cal
ΔU = -7677 cal
Problem 31
In the study of the isothermal compression of an ideal gas, measurements are first taken at 42°C and then at 27°C. In the last measurement, this product is found to increase by:
c) 5%
T1 = 300 K
T2 = 315 K
300 → 100
315 → xx = 105 – 100 = 5%
Problem 32
Heat is added to oxygen at constant pressure from 273°C to 373°C, performing 20 J of work. This means the mass of oxygen is:
a) 0.77 g