The First Law of Thermodynamics
20-1. In an industrial chemical process, 600 J of heat is supplied to a system while 200 J of work is done BY the system. What is the increase in the internal energy of the system?
Work done BY the system is positive, heat INTO a system is positive. Apply first law:
ΔQ = ΔU + ΔW; ΔQ = 600 J; ΔW = 200 J
ΔU =ΔQ - ΔW = 600 J – 200 J; ΔU = 400 J
20-2. Assume that the internal energy of a system decreases by 300 J while 200 J of work is done bv a gas. What is the value of Q? Is heat lost or gained by the system?
ΔU = -300 J; ΔW = +200 J; ΔQ = ΔQ + ΔW
ΔQ = (-300 J) + (200 J) = - 100 J; Heat Lost: ΔQ = -100 J
20-3. In a thermodynamic process, the internal energy of the systemincreases by 500 J. How much work was done by the gas if 800 J of heat is absorbed?
ΔU = +500 J; ΔQ = +800 J; ΔQ = ΔU + ΔW
ΔW = ΔQ - ΔU = 800 J – 500 J; ΔW = 300 J
20-4. A piston does 3000 ft lb of work on a gas, which then expands performing 2500 ft lb of work on its surroundings. What is the change in internal energy of the system if net heat exchange is zero? [ Assume ΔQ =0, then ΔQ = ΔU + ΔW = 0 and ΔU = -ΔW ]
ΔU = -(Workout – Workin) = -(2500 ft lb – 3000 ft lb); ΔU = + 500 ft lb = 0.643 Btu
20-5. In a chemical laboratory, a technician applies 340 J of energy to a gas while the system surrounding the gas does 140 J of work ON the gas. What is the change in internal energy? [ ΔQ = +340 J; ΔW = -140 J (work ON gas is negative) ]
ΔU = ΔQ - ΔW =(350 J) – (-140 J); ΔU = 480 J
20-6. What is the change in internal energy for Problem 20-5 if the 140 J of work is done BY the gas instead of ON the gas? [ Work BY gas is positive, ΔW = +140 J ]
ΔU = ΔQ - ΔW = (340 J) – (+140 J); ΔU = 200 J
20-7. A system absorbs 200 J of heat as the internal energy increases by 150 J. What work is done bv the gas? [ ΔQ = +200 J, ΔU = +150 J ]ΔW = ΔQ - ΔU = 200 J – 100 J = -50 J; ΔW = 50 J
*20-8. The specific heat of water is 4186 J/kg C0. How much does the internal energy of 200 g of water change as it is heated from 200C to 300C? Assume the volume is constant.
ΔQ = mcΔT = (0.2 kg)(4186 J/kg)(300C – 200C); ΔQ = 8372 J
Since ΔV = 0, ΔW is also zero: ΔU = ΔQ: ΔU = +8370 J
*20-9. At a constant pressureof 101.3 kPA, one gram of water (I cm3) is vaporized completely and has a final volume of 1671 cm3 in its vapor form. What work is done by the system against its surroundings? What is the increase in internal energy? (1 cm3 = 1 x 10-6 m3) Work = PΔV = (101,300 Pa)(1671 cm3 – 1cm3)(10-6 m3/cm3); ΔW = 169 J
*20-9. (Cont.) ΔQ = mLf = (0.001 kg)(2.256 x 106 J/kg) = 2256 J
ΔU =ΔQ - ΔW = 2256 J – 169 J; ΔU = 2090 J
*20-10. A I0-kg block slides down a plane from a height of I0 in, and has a velocity of I0 m/s when it reaches the bottom, how many calories of heat were lost due to friction?
20-11. An ideal gas expands isothermally while absorbing 4.80 J of heat. The piston has a mass of 3 kg. How high will the piston rise above itsinitial position? [ ΔU = 0 (isothermal) ]
ΔQ = ΔU + ΔW; ΔW = ΔQ = +4.80 J; Work = Fh = 4.80 J
[pic]; h = 0.163 m or 16.3 cm
20-12. The work done on a gas during an adiabatic compression is 140 J. Calculate the increase in internal energy of the system in calories.
For an adiabatic process, ΔQ = 0 and work ON gas is ΔW = -140 J
ΔU + ΔW = 0; ΔU = -ΔW = -(-140 J); ΔU =+140 J
The internal energy increases as work is done in compressing the gas.
20-13. During an isobaric expansion a steady pressure of 200 kPa causes the volume of a gas to change from I L to 3 L. What work is done by the gas? [ 1 L = 1 x 10-3 m3 ]
Work = P(Vf – Vi) = (200,000 Pa)(3 x 10-3 m3 – 1 x 10-3 m3)
Work = 400 J
20-14. A gas is confined to a copper can. How much heat...