Physics thermodynamic

Hi!

This is a tough one:

  • using PV=nRT (with PnR constant in this question), we can see V is directly proportional to T, so if V * 1,9 then T also * 1,9
  • Ti=10°C=283°K
  • Tf=10 * 1,9=19°C=292°K
  • n=m/M so n(He)=10000
  • n=V(in dm³)/22,4 so Vi=22,4 * 10⁴
  • Vf=Vi * 1,9=42,56 * 10⁴
  • deltaU=Q-W
    = mc(deltaT) - (P*(deltaV))
    = (10 * 10⁴ * 5190 * (292-283)) - (1 * (42,56 * 10⁴ - 22,4 * 10⁴))
    = 47 * 10⁷ - 20,16 * 10⁴ atm/L
  • 1atm/L=101,3J so 20,16 * 10⁴=20,4 * 10⁶J
    so deltaU = 47 * 10⁷ - 20,4 * 10⁶
    =45,04 * 10⁷J
    =0,4504MJ
    = about 0,5MJ

Hope this helps!

why you’ve multiplied 20,16*10^4 by 101,3 ?

Hi! Because we need the unit of Work to be in Joules not in atm/L