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Chapters 20 and 21 problems. 001(part 1 of 1) 0 points
A 0.0505 kg ingot of metal is heated to 173◦C and then is dropped into a beaker containing 0.426 kg of water initially at 23◦C.
If the final equilibrium state of the mixed system is 25.4◦C, find the specific heat of the metal. The specific heat of water is 4186 J/kg · ◦C. Correct answer: 574.172 J/kg · ◦C. Explanation: Given : mw = 0.426 kg , mx= 0.0505 kg , cw = 4186 J/kg · ◦C , Tw = 23◦C , Ti= 173◦C , and Tf = 25.4◦C .
Because the thermal energy lost by the in-got equals the thermal energy gained by the water, mxcx(Ti−Tf) = mwcw(Tf −Tw) , cx = mwcw(Tf −Tw) mx(Ti−Tf) = (0.426 kg) (4186 J/kg · ◦C) 0.0505 kg × 25.4 ◦C − 23◦C 173◦C − 25.4◦C = 574.172 J/kg · ◦C . 002(part 1 of 1) 0 points
337 g of water at 70◦C is poured into a 25 g aluminum cup containing 66 g of water at 29◦C.
What is the equilibrium temperature of the system? Assume the specific heat of alu-minum is 0.215 cal/g ·◦C. Correct answer: 62.8341◦C. Explanation: Given : mh = 337 g , mc = 66 g , mAl = 25 g , cw = 1 cal/g ·◦C , cAl = 0.215 cal/g ·◦C , Tc = 29◦C , and Th = 70◦C .
The heat lost by hot water equals the heat gained by cold metal plus the water.
mhcw(Th−Tf) = mAlcAl(Tf −Tc) + mccw(Tf −Tc) (mhcw+ mAlcAl+ mccw) Tf = mhcwTf + mAlcAlTc+ mccwTc Tf = mhcwTh+ mAlcAlTc+ mccwTc mhcw+ mAlcAl+ mccw Since mhcwTh+ mAlcAlTc+ mccwTc = (337 g) (1 cal/g ·◦C) (70◦C) + (25 g) (0.215 cal/g ·◦C) (29◦C) + (66 g) (1 cal/g ·◦C) (29◦C) = 25659.9 cal and mhcw+ mAlcAl+ mccw = (337 g) (1 cal/g ·◦C) + (25 g) (0.215 cal/g ·◦C) + (66 g) (1 cal/g ·◦C) = 408.375 cal/◦C ,
then the equilibrium temperature will be
Tf =
25659.9 cal 408.375 cal/◦C
= 62.8341◦C ,
003(part 1 of 1) 10 points
A calorimeter contains 410 mL of water at 48◦C and 43 g of ice at 0◦C.
Find the final temperature of the system. Correct answer: 35.8923 ◦C.
Given : V = 410 mL , mw = 410 g , mice= 0.043 kg , cw = 1 cal/g ·◦C , Lf = 333000 J/kg , Tw = 48◦C , Ti= 0◦C , and ∆T = 48◦C .
The 410 mL = mv of water cools to 0◦C,
giving off
Qw = mwcw∆T
= (410 g) (1 cal/g ·◦C) (48◦C) = 19680 cal
heat.
If all of the ice melts, it absorbs Qi= miceLf
= (0.043 kg) (333000 J/kg) (0.2389 cal/J) = 3420.81 cal
heat.
Since Qw > Qi, all of the ice melts and also
warms up to a temperature above 0◦C: Thus Qlostwater = Qmeltice+ Qgained
mwcw( Tw−Tf) = Qi+ micecw( Tf −Tice) mwcwTw−mwcwTf = Qi+ micecw( Tf −0) mwcwTw−Qi = ( micecw + mwcw)Tf Tf = mwcwTw−Qi micecw+ mwcw = (410 g)(1 cal/g · ◦C)(48◦C) − 3420.81 cal (0.043 kg)(1 cal/g ·◦C) + (410 g)(1 cal/g ·◦C) = 19680 cal − 3420.81 cal 43 cal/◦C + 410 cal/◦C = 35.8923◦C . 004(part 1 of 3) 0 points
4 cm3 of water is boiled at atmospheric pres-sure to become 3630.5 cm3 of steam, also at atmospheric pressure.
Calculate the work done by the gas during this process. Correct answer: 367.364 J. Explanation: Given : P = 101300 Pa , V1= 4 cm3, V2= 3630.5 cm3, m = 0.004 kg , and Lv = 2.26 × 106 J/kg . W = P ∆V = P (V2−V1) = (101300 Pa)(3630.5 cm3−4 cm3) · µ 1 m 100 cm ¶3 = 367.364 J. 005(part 2 of 3) 0 points
Find the amount of heat added to the water to accomplish this process.
Correct answer: 9040 J. Explanation: Q = m Lv = (0.004 kg)(2.26 × 106 J/kg) = 9040 J. 006(part 3 of 3) 0 points Find the change in internal energy. Correct answer: 8672.63 J.
Explanation:
∆U = Q − W = 9040 J − 367.364 J = 8672.63 J.
007(part 1 of 3) 4 points
Two moles of helium gas initially at 252 K and 0.33 atm are compressed isothermally to 1.57 atm.
Find the final volume of the gas. Assume the helium to behave as an ideal gas.
Correct answer: 0.0263486 m3. Explanation: Given : n = 2 mol , R = 8.31451 J/K · mol , Tf = 252 K , and Pf = 1.57 atm .
From the ideal gas law, PfVf = n R T Vf = n R T Pf = (2 mol)(8.31451 J/K · mol)(252 K) (1.57 atm)(101300 Pa/atm) = 0.0263486 m3. 008(part 2 of 3) 3 points Find the work done by the gas. Correct answer: −6.5361 kJ. Explanation:
Given : Pi= 0.33 atm and
Ti= 252 K .
The initial volume was Vi = n R T Pi = (2 mol)(8.31451 J/K · mol)(252 K) (0.33 atm)(101300 Pa/atm) = 0.125356 m3
and the work done was W = Z P dV = n R T ln µ Vf Vi ¶ = (2 mol)(8.31451 J/K · mol)(252 K) lnµ 0.0263486 m 3 0.125356 m3 ¶ 1 kJ 1000 J = −6.5361 kJ . 009(part 3 of 3) 3 points Find the thermal energy transferred. Correct answer: −6.5361 kJ.
Explanation:
From the first law of thermodynamics, since ∆U = 0,
Q = W
= −6.5361 kJ .
010(part 1 of 2) 5 points
A gas is taken through the cyclic process de-scribed by the figure. Let a = 9, b = 9.
2 4 6 8 10 12 14 2 4 6 8 10 12 14 A B C V (m3) P(kPa)
Find the net energy transferred to the sys-tem by heat during one complete cycle. Correct answer: 40.5 kJ.
Explanation:
Given : ∆P = 9000 Pa and ∆V = 9 m3. The change in internal energy is
∆Ucycle= Qcycle+ Wcycle = 0
Qcycle = −WABCA
= (area enclosed in P V diagram) = 1 2¡9 m 3 ¢ (9000 Pa) = 40500 J = 40.5 kJ . Alternate Solution:
For each step the work is the negative of the area under the curve on the PV diagram:
WAB = −[(3.6 kPa)¡10.5 m 3 −6.5 m3¢ +1 2(9.6 kPa − 3.6 kPa) ·¡10.5 m3−6.5 m3¢] = −26.4 kJ WBC = 0 kJ WCA = −(3.6 kPa)¡6.5 m3−10.5 m3¢ = 14.4 kJ Qcycle = −WABC = −(−26.4 kJ + 0 kJ + 14.4 kJ) = 12 kJ . 011(part 2 of 2) 5 points
If the cycle is reversed (that is, the process follows the path ACBA), what is the net en-ergy transferred by heat per cycle?
Correct answer: −40.5 kJ. Explanation:
If the cycle is reversed, then
Qcycle= WACBA = −WABCA= −40.5 kJ .
012(part 1 of 2) 0 points
An ideal gas initially at 333 K undergoes an isobaric expansion at 2 kPa.
If the volume increases from 1.6 m3 to 4 m3 and 15.9 kJ of thermal energy is transferred to the gas, find the change in its internal energy. Correct answer: 11.1 kJ. Explanation: Given : P = 2 kPa , Vf = 4 m 3 , Vi = 1.6 m3, and Ti = 333 K .
From the first law of thermodynamics ∆U = Q − W = Q − P (Vf −Vi)
= 15.9 kJ − (2 kPa)(4 m3−1.6 m3) = 11.1 kJ .
013(part 2 of 2) 0 points Find the final temperature of the gas. Correct answer: 832.5 K.
Explanation:
From the ideal gas law Vi Ti = Vf Tf Tf = VfTi Vi = (4 m 3 )(333 K) (1.6 m3 ) = 832.5 K . 014(part 1 of 2) 0 points
One mole of an ideal gas does 2519 J of work on the surroundings as it expands isother-mally to a final pressure of 0.8 atm and vol-ume of 33 L.
Determine the initial volume. (R = 8.31451 J/K · mol.) Correct answer: 12.866 L. Explanation: Given : n = 1 mol , W = 2519 J , R = 8.31451 J/K · mol , Pf = 0.8 atm , and Vf = 33 L .
By the ideal gas law, P V = n R T . The work done in an isothermal process is
W = n R T ln µ Vf Vi ¶ = P V ln µ Vf Vi ¶ ln µ Vf Vi ¶ = W P V Vf Vi = exp· W P V ¸ Vi = Vfexp · −W P V ¸ Since
W P V = 2519 J (0.8 atm)(101300)(33 L)(1e − 3) = 0.941922 Vi = 33 L exp [−0.941922] = 12.866 L . 015(part 2 of 2) 0 points Determine the temperature of the gas. Correct answer: 321.645 K.
Explanation:
According to the equation of state for an ideal gas, the temperature is
Tf = PfVf n R = (0.8 atm)(101300)(33 L)(1e − 3) (1 mol)(8.31451 J/K · mol) = 321.645 K . 016(part 1 of 2) 0 points
Air in the cylinder of a diesel engine at 21.8◦C is compressed from an initial pres-sure of 0.821 atm and of volume of 793 cm3 to a volume of 86 cm3.
Assuming that air behaves as an ideal gas (γ = 1.40) and that the compression is adia-batic and reversible, find the final pressure. Correct answer: 18.4089 atm.
Explanation:
Given : Pi= 0.821 atm ,
Vi= 793 cm3, and
Vf = 86 cm3.
Using the relation
PiViγ = PfVfγ, we find that Pf = Pi µ Vi Vf ¶γ = (0.821 atm ) µ 793 cm 3 86 cm3 ¶1.40 = 18.4089 atm . 017(part 2 of 2) 0 points
Find the final temperature under the same assumptions as above. Correct answer: 443.865◦C. Explanation: Given : Ti= 21.8◦C = 294.8 K . Since P V = n R T
is always valid during the process and since no gas escapes from the cylinder,
PiVi Ti = PfVf Tf = n R . Therefore, Tf = PfVf PiVi Ti = (18.4089 atm) (86 cm 3 ) (0.821 atm) (793 cm3) (294.8 K) = 716.865 K = 443.865◦C . 018(part 1 of 4) 3 points
4.2 L of diatomic gas (γ = 1.4) confined to a cylinder are put through a closed cycle. The gas is initially at 1 atm and at 330 K. First, its pressure is tripled under constant volume. Then it expands adiabatically to its original pressure and finally is compressed isobarically to its original volume.
Determine the volume at the end of the adiabatic expansion. Correct answer: 9.20556 L. Explanation: Given : V0= 4.2 L = 0.0042 m3 and Pf = 3 Pi. Basic Concepts P V = n R T ∆U = Q − Wgas U = n CvT
Q = n C ∆T
Solution: For an adiabatic process we have PBV γ B = PCV γ C , so 3 P0V0γ = P0VCγ.
It follows from this that
VC = V0 µ 3 P0 P0 ¶ 1 γ = (4.2 L) (3) 1 1.4 = 9.20556 L . 019(part 2 of 4) 3 points
Find the temperature of the gas at the start of the adiabatic expansion.
Correct answer: 990 K. Explanation:
Given : T0 = 330 K .
From the equation of state for an ideal gas we obtain PBVB = n R TB = 3 P0V0 = 3 n R T0 Therefore TB = 3 T0 = 3 (330 K) = 990 K . 020(part 3 of 4) 2 points
Find the temperature at the end of the adia-batic expansion. Correct answer: 723.294 K. Explanation: TB V γ−1 B = TC V γ−1 C TC = TB µ VB VC ¶γ−1 = (990 K) µ 4.2 L 9.20556 L ¶1.4−1 = 723.294 K . 021(part 4 of 4) 2 points What is the net work done for this cycle? Correct answer: 352.579 J.
Explanation:
Given : P0 = 1 atm = 101300 Pa .
Since for a closed cycle ∆U = 0, net work done for this cycle equals net heat transferred to the gas. In AB this heat is
QAB = n Cv∆T = 5 2n R (TB −TA) = 5 2n R (3 T0−T0) = 5 n R T0 = 5 P0V0 = 5 (101300 Pa) (0.0042 m3) = 2127.3 J . In BC QBC = 0,
since the process is adiabatic, and with the help of the equation of the state for an ideal gas, we have QCA = n Cp∆T = 7 2n R (TA −TC) = 7 2n R µ T0−T0VC V0 ¶ = 7 2P0V0 µ 1 −VC V0 ¶ = 7 2(101300 Pa) (0.0042 m 3 ) × µ 1 −9.20556 L 4.2 L ¶ = −1774.72 J .
For the whole cycle QABCA = QAB+ QCA
= (2127.3 J) + (−1774.72 J) = 352.579 J .
022(part 1 of 2) 0 points
A cylinder contains 2.46 mol of helium gas at a temperature of 285 K.
How much heat must be transferred to the gas to increase its temperature to 602 K if it is heated at constant volume? The molar specific heat at constant volume, cv, of helium
is 12.5 J/mol/K. Correct answer: 9747.75 J. Explanation: Given : n = 2.46 mol , Ti = 285 K , and Tf = 602 K .
For the constant-volume process, the work done is zero. Therefore for the heat Q1
trans-ferred to the gas during the process we have Q1 = ∆U = 3 2n R ∆T = n cv(Tf −Ti) , Q1= (2.46 mol) (12.5 J/mol/K) ·(602 K − 285 K) = 9747.75 J . 023(part 2 of 2) 0 points
How much thermal energy must be trans-ferred to the gas at constant pressure to raise the temperature to 602 K? The molar spe-cific heat at constant presure, cp, of helium is
20.8 J/mol/K.
Correct answer: 16220.3 J. Explanation:
For the thermal energy Q2 transferred to
the gas at constant pressure, Q2 = n cp∆T
= (2.46 mol) (20.8 J/mol/K) (317 K) = 16220.3 J .
024(part 1 of 3) 0 points
Two moles of an idea gas (γ = 1.4) expands
slowly and adiabatically from a pressure of 4 atm and a volume of 13 L to a final volume of 33 L.
What is the final pressure? Correct answer: 1.08558 atm. Explanation: Given : P0= 4 atm = 405200 Pa , V0= 13 L = 0.013 m3, and V1= 33 L = 0.033 m3. Basic Concepts P Vγ = const P V = n R T
Solution: Since an adiabatic process for an ideal gas is described by P Vγ = const, we have P0V0γ = P1V1γ, so P1 = P0 µ V0 V1 ¶γ = (4 atm)µ 13 L 33 L ¶1.4 = 1.08558 atm . 025(part 2 of 3) 0 points What is the initial temperature? Correct answer: 316.772 K. Explanation:
From the equation of state for an ideal gas we have T0 = P 0V0 n R = (405200 Pa) (0.013 m 3 ) (2 mol) (8.31451 J/K · mol) = 316.772 K . 026(part 3 of 3) 0 points What is the final temperature? Correct answer: 218.232 K. Explanation:
From the equation of state for an ideal gas we have T1 = P 1V1 n R = (109969 Pa) (0.033 m 3 ) (2 mol) (8.31451 J/K · mol) = 218.232 K . 027(part 1 of 4) 0 points
Assume that a molecule has f degrees of freedom. Consider a gas consisting of such molecules.
Determine its total thermal energy. 1.U = f 3 n R T 2 2.U = f n R T 3 3.U = f 2 n R T 3 4.U = f n R T 2 correct 5.U = f n R T Explanation: Cv = 1 n µ dU dT ¶
According to the equipartition theorem, the amount of energy per each degree of freedom is kBT 2 , so that U = N f kBT 2 = f n R T 2 . 028(part 2 of 4) 0 points
Find its molar specific heat at constant vol-ume. 1.Cv = f R 2.Cv = 1 2f R correct 3.Cv = 2 3f R 4.Cv = 1 4f R 5.Cv = 1 3f R Explanation:
The specific heat is given by Cv = 1 n µ dU dT ¶ = 1 2f R . 029(part 3 of 4) 0 points
Find its molar specific heat at constant pres-sure. 1.Cp = 1 2(f + 3)R 2.Cp = 3 2(f + 2) R 3.Cp = 1 2(f + 1) R 4.Cp = 1 3(f + 2) R 5.Cp = 1 2(f + 2) R correct Explanation: Since Cp = Cv+ R, we have Cp= Cv+ R = 1 2(f + 2) R . 030(part 4 of 4) 0 points Determine the ratio γ = Cp
Cv 1.γ = f + 2 2 f 2.γ = f + 2 3 f 3.γ = f + 2 f correct 4.γ = f + 3 f 5.γ = f + 1 f Explanation:
Taking the magnitudes of Cp and Cv from
the previous sections, we obtain γ = Cp
Cv =
f + 2 f .
