The final temperature and initial temperature of a diatomic ideal gas are Tƒ and To, respectively. The change in entropy in an isobaric process is given by © C 5 -nRln 3 -nRln Tf To Tf To Tf
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- In an isothermal reversible expansion at 27 , an ideal gas does 20 J of work. What is the entropy change of the gas?A copper rod of cross-sectional area 5.0 cm2 and length 5.0 m conducts heat from a heat reservoir at 373 K to one at 273 K. What is the time rate of change of the universe's entropy for this process?One mole of an ideal monatomic gas is confined to a rigid container. When heat is added reversibly to die gas, its temperature changes from T1 to T2 . (a) How much heat is added? (b) What is the change in entropy of the gas?
- On an adiabatic process of an ideal gas pressure, volume and temperature change such that pV is constant with =5/3 for monatomic gas such as helium and =7/5 for diatomic gas such as hydrogen at room temperature. Use numerical values to plot two isotherms of 1 mol of helium gas using ideal gas law and two adiabatic processes mediating between them. Use T1=500K,V1=1L, and T2=300K for your plot.Two moles of a monatomic ideal gas such as oxygen is compressed adiabatically and reversibly from a state (3 atm, 5 L) to a state with a pressure of 4 atm. (a) Find the volume and temperature of the final state. (b) Find the temperature of the initial state. (c) Find work done by the gas in the process. (d) Find the change in internal energy in the process. Assume Cv=5R and Cp=Cv+R for the diatomic ideal gas in the conditions given.In an isochoric process, heat is added to 10 mol of monoatomic ideal gas whose temperature increases from 273 to 373 K. What is the entropy change of the gas?
- (a) A 5.0-kg rock at a temperature of 20 is dropped into a shallow lake also at 20 from a height of 1.0103 m. What is the resulting change in entropy of the universe? (b) If the temperature of the lock is 100 when it is dropped, what is the change of entropy of the universe? Assume that air friction is negligible (not a good assumption) and that c=860 J/kg K is the specific heat of the rock.After a free expansion to quadruple its volume, a mole of ideal diatomic gas is compressed back to its original volume isobarically and then cooled down to its original temperature. What is the minimum heat removed from the gas in the final step to restoring its state?A cylinder containing three moles of a monatomic ideal gas is heated at a constant pressure of 2 atm. The temperature of the gas changes from 300 K to 350 K as a result of the expansion. Find work done (a) on the gas; and (b) by 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 ale as follows: i. Isobaric intake stroke (OA). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure P0 as the piston expands, increasing the volume of the cylinder from zero to VA . ii. Adiabatic compression stroke (AB). The temperature of the mixture rises as the piston compresses it adiabatically from a volume VA to VB . iii. 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 Q1 causes the pressure to increase from pB to pc at the constant volume VB(=Vc) . iv. Adiabatic expansion (CD). The heated mixture of gasoline and air expands against the piston, increasing the volume from VC to VD . This is called the power stroke, as it is the part of the cycle that delivers most of the power to the crankshaft. v. 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) . Most of the available energy is lost here, as represented by the heat exhaust Q2 . vi. Isobaric compression (AO). The exhaust valve remains open, and the compression from VA to zero drives out the remaining combustion products. (a). Using (i)e=W/Q1; (ii)w=Q1Q2; and (iii)Q1=nCv(TCTB),Q2=nCv(TDTA), Show that e=1TDTATCTB. (b). Use the fact that steps (ii) and (iv) are adiabatic to show that e=11r1 where r=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=6 and =1.4 (the value for air), e=0.51 , or an efficiency of 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% to 30%.An ideal gas expands quasi-statically to three times its original volume. Which process requires more work from the gas, an isothermal process or an isobaric one? Determine the ratio of the work done in processes.The energy output of a heat pump is greater than the energy used to operate the pump. Why doesn't this statement violate the first law of thermodynamics?