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University Physics Volume 2

Challenge Problems

University Physics Volume 2Challenge Problems

Challenge Problems


(a) An infinitesimal amount of heat is added reversibly to a system. By combining the first and second laws, show that dU=TdSdWdU=TdSdW. (b) When heat is added to an ideal gas, its temperature and volume change from T1andV1toT2andV2T1andV1toT2andV2. Show that the entropy change of n moles of the gas is given by



Using the result of the preceding problem, show that for an ideal gas undergoing an adiabatic process, TVγ1TVγ1 is constant.


With the help of the two preceding problems, show that ΔSΔS between states 1 and 2 of n moles an ideal gas is given by



A cylinder contains 500 g of helium at 120 atm and 20°C20°C. The valve is leaky, and all the gas slowly escapes isothermally into the atmosphere. Use the results of the preceding problem to determine the resulting change in entropy of the universe.


A diatomic ideal gas is brought from an initial equilibrium state at p1=0.50atmp1=0.50atm and T1=300KT1=300K to a final stage with p2=0.20atmp2=0.20atm and T1=500K.T1=500K. Use the results of the previous problem to determine the entropy change per mole of the gas.


The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle, which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle are as follows:

  1. Isobaric intake stroke (OA). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure p0p0 as the piston expands, increasing the volume of the cylinder from zero to VAVA.
  2. Adiabatic compression stroke (AB). The temperature of the mixture rises as the piston compresses it adiabatically from a volume VAtoVBVAtoVB.
  3. Ignition at constant volume (BC). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat Q1Q1 causes the pressure to increase from pBtopCpBtopC at the constant volume VB(=VC)VB(=VC).
  4. Adiabatic expansion (CD). The heated mixture of gasoline and air expands against the piston, increasing the volume from VCtoVDVCtoVD. This is called the power stroke, as it is the part of the cycle that delivers most of the power to the crankshaft.
  5. Constant-volume exhaust (DA). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at VA(=VD)VA(=VD). Most of the available energy is lost here, as represented by the heat exhaust Q2Q2.
  6. Isobaric compression (AO). The exhaust valve remains open, and the compression from VAVA to zero drives out the remaining combustion products.

(a) Using (i) e=W/Q1e=W/Q1; (ii) W=Q1Q2W=Q1Q2; and (iii) Q1=nCv(TCTB)Q1=nCv(TCTB), Q2=nCv(TDTA)Q2=nCv(TDTA), show that


(b) Use the fact that steps (ii) and (iv) are adiabatic to show that


where r=VA/VBr=VA/VB. The quantity r is called the compression ratio of the engine.

(c) In practice, r is kept less than around 7. For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for r=6r=6 and γ=1.4γ=1.4 (the value for air), e=0.51e=0.51, or an efficiency of 51%.51%. Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about 25%to30%25%to30%.

The figure shows a closed loop graph with four points A, B, C and D. The x-axis is V and y-axis is p. The value of V at A and D is equal and at B and C is equal.

An ideal diesel cycle is shown below. This cycle consists of five strokes. In this case, only air is drawn into the chamber during the intake stroke OA. The air is then compressed adiabatically from state A to state B, raising its temperature high enough so that when fuel is added during the power stroke BC, it ignites. After ignition ends at C, there is a further adiabatic power stroke CD. Finally, there is an exhaust at constant volume as the pressure drops from pDpD to pApA, followed by a further exhaust when the piston compresses the chamber volume to zero.

(a) Use W=Q1Q2W=Q1Q2, Q1=nCp(TCTB)Q1=nCp(TCTB), and Q2=nCv(TDTA)Q2=nCv(TDTA) to show that e=WQ1=1TDTAγ(TCTB)e=WQ1=1TDTAγ(TCTB).

(b) Use the fact that ABAB and CDCD are adiabatic to show that


(c) Since there is no preignition (remember, the chamber does not contain any fuel during the compression), the compression ratio can be larger than that for a gasoline engine. Typically, VA/VB=15andVD/VC=5VA/VB=15andVD/VC=5. For these values and γ=1.4,γ=1.4, show that ε=0.56ε=0.56, or an efficiency of 56%56%. Diesel engines actually operate at an efficiency of about 30%to35%30%to35% compared with 25%to30%25%to30% for gasoline engines.

The figure shows a closed loop graph with four points A, B, C and D. The x-axis is V and y-axis is p. The value of V at A and D is equal and the value of p at B and C is equal.

Consider an ideal gas Joule cycle, also called the Brayton cycle, shown below. Find the formula for efficiency of the engine using this cycle in terms of P1P1, P2P2, and γγ.

The figure shows a closed loop graph with four points 1, 2, 3 and 4. The x-axis is V and y-axis is p. The value of p at 1 and 4 is equal and at 2 and 3 is equal

Derive a formula for the coefficient of performance of a refrigerator using an ideal gas as a working substance operating in the cycle shown below in terms of the properties of the three states labeled 1, 2, and 3.

The figure shows a closed loop graph with three points 1, 2 and 3. The x-axis is V and y-axis is p. The value of V at 1 and 2 is equal and the value of p at 2 and 3 is equal.

Two moles of nitrogen gas, with γ=7/5γ=7/5 for ideal diatomic gases, occupies a volume of 10−2m310−2m3 in an insulated cylinder at temperature 300 K. The gas is adiabatically and reversibly compressed to a volume of 5 L. The piston of the cylinder is locked in its place, and the insulation around the cylinder is removed. The heat-conducting cylinder is then placed in a 300-K bath. Heat from the compressed gas leaves the gas, and the temperature of the gas becomes 300 K again. The gas is then slowly expanded at the fixed temperature 300 K until the volume of the gas becomes 10−2m310−2m3, thus making a complete cycle for the gas. For the entire cycle, calculate (a) the work done by the gas, (b) the heat into or out of the gas, (c) the change in the internal energy of the gas, and (d) the change in entropy of the gas.


A Carnot refrigerator, working between 0°C0°C and 30°C30°C is used to cool a bucket of water containing 10−2m310−2m3 of water at 30°C30°C to 5°C5°C in 2 hours. Find the total amount of work needed.

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