(a) For this problem, we assume that the gas obeys the Ideal Gas law. Since it is a constant volume, the Ideal Gas law states that P/T = nR/V = constant. For this gas, we get P/T = .7 atm/373 K = .00187 atm/K. Thus, if the pressure is 0.0400 atm, then T = 0.0400 atm/(.00187 atm/K) = 21.3 K.
(b) For T = 723 K, we get P = (.00187 atm/K)(723 K) = 1.35 atm.
a) From the Ideal Gas law, we know that PV = nRT. If the volume is held constant (Vo = Vf) and the pressure triples (3Po = Pf), then PoVo/To = PfVf/Tf becomes PoVo/To = 3PoVo/Tf. This gives Tf = 3To = 3(300 K) = 900 K.
b) If both the pressure and volume are doubled, then PfVf/Tf = 4PoVo/Tf. From the Ideal Gas law, this gives Tf = 4To = 4(300 K) = 1200 K.
If we assume that all of the heat leaving the horseshoe enters the water, then we have mironciron(500 oC - Tf) = mwatercwater(Tf - 22 oC). Solving this for Tf, we get Tf = (mironciron 500 oC + mwatercwater 22 oC)/(mironciron + mwatercwater) = ((.4 kg)(448 J/kg oC)(500 oC) + (20 kg)(4186 J/kg oC)(22 oC))/((.4 kg)(448 J/kg oC) + (20 kg)(4186 J/kg oC)) = (89600 J + 1842000 J)/(179 J/oC + 83700 J/oC) = 23.0 oC.
We know that the maximum efficiency that a heat engine can have is given by ECarnot = 1 - Tc/TH. For the temperatures given, this means ECarnot = 1 - 100 K/200 K = 1 - .50 = .50. This means that the maximum efficiency is 50%, which means that the student could not have an efficiency of 60%.