Question

A balloon filled with 39.1 moles of helium gas has a volume of 876 L at 0.0°C and 1.00 atm pressure. At constant pressure, the temperature of the balloon is increased to 38.0°C causing the balloon to expand. Calculate q, w, ∆E for the helium in the balloon. The specific heat of helium gas is 5.20 J/(g °C) and 101.3 J = 1 Latm.

Answer 1

Vi = 876 L
Ti= 0 oC = 273 K
Tf = 38 oC = (273+38) K = 311 K
P = 1 atm.

use:
Vf/Vi = Tf/Ti
Vf/876 = 311/273
Vf= 998 L

FIND W:
W = -P(Vf-Vi)
     = -1 atm (998 L - 876 L)
     = -122 atm L
      = -122 * 101.3 J   {Since 1 atm-L= 101.3 J}
      = -12358.6 J

mass of helium gas = number of moles *molar mass
                                        = 39.1 * 4
                                        = 156.4 g

FIND ∆E
∆E =m*CV*(Tf-Ti)
        = 156.4 * 5.2* (311-273)
        =30904.64 J

Find Q:
use:
∆E = Q + W
30904.64 J = Q + (-12358.6 J)
Q =43263.24 J