跪求化學高手解此普化題目....

A balloonist is preparing to make a trip in a helium-filled balloon. The trip begins in early morning at a temperature of 15度C. By midafternoon

the temperature has increased to 30度C. Assuming the pressure remains constant at 1.00 atm

for each mole of helium calculate:(a)the initial and final volumes (ans: initial=23.65L final=24.89L)(b)the change in internal energy.{Hint: helium behaves like an ideal gas. So E=3/2nRT.Note the unit of R are consistent with those of E.} (ans: 187J)(c)The work(w) done by the helium{in J} (ans:-120J)(d)The heat(q) transferred{in J} (ans:310J)P.S. 以上題目附了答案...原因是希望各位化學高手可以將過程告訴敝人...尤其是(c)(d)兩題更是不解...希望各位大大可以幫幫忙!!
(a) 假設氦為理想氣體

V=nRT/P

R=0.08206 atm.L/mol.oK

Vi=0.08206*(273.15 15)=23.65 L

Vf=0.08206*(273.15 30)=24.88 L.(b) dE=(3/2)nRdT

R=8.31431 J/mol.oK

dE=1.5*8.31431*15=187 J(c) 1 atm.L = 10333.23 kg.L/m2 * 9.8 m/s2 * 0.001 m3/L = 101.3 kg.m2/s2 = 101.3 J

w=PdV=23.65-24.88=-1.23 atm.L=-1.23*101.3 J=-125 J(d) q=dE-w=187-(-125)=312 J
(a)如題

定壓下溫度從15度升到30度由PV=nRT可個別求出其初體積與末體積1xV(初)=1x0.082x(273 15)

V(初)=23.616L(因各數值有效位數不同

故與解答不盡相同)同理得V(末)=24.846L(b)能量變化=末能減初能=3/2nR(T末-T初)=3/2x1x8.314x15=187.065J(c)W=-[P(末)V(末)-P(初)V(初)]=-[1x24.846-1x23.616]=-1.23J(d)E=q W

故q=187.065 1.23=188.353(J)

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