Question

A balloon filled with 39.1 moles of helium has a volume of 876 l at 0.0'c and 1.00 atm pressure. the temperature of the balloon is increased to 38.0'c as it expands to a volume of 998 l, the pressure remaining constant. calculate q, w, and #e for the helium in the balloon. (the molar heat capacity for helium gas is 20.8 j/°c # mol.)

Answer 1

Answer:

Q = 30.90 kJ, W = 12.36 kJ, ΔE = 18.54 kJ.

Explanation:

The expansion is happening at constant pressure without a phase change, thus, the heat can be calculated by:

Q = n*c*ΔT

Where n is the number of moles, c is the molar heat capacity, and ΔT is the temperature variation (final - initial), thus:

Q = 39.1*20.8*(38.0 - 0.0)

Q = 30904.64 J

Q = 30.90 kJ

The work done by a expansion at constant pressure is:

W = P*ΔV

Where P is the pressure (1 atm = 101325 Pa), and ΔV the volume variation (final - initial). Vfinal = 998 L = 0.998m³, Vinitial = 876 L = 0.876 m³.

W = 101325*(0.998 - 0.876)

W = 12361.65 J

W = 12.36 kJ

By the first law of the thermodynamics, the variation of the internal energy ΔE is:

ΔE = Q - W

ΔE = 30.90 - 12.36

ΔE = 18.54 kJ

Answer 2

Moles of Helium = 39.1 moles

ΔT = (38.08 – 0.08) ֯C = 38.00 ֯C

Q = nC ΔT = 39.1 * 20.8 J/ ֯C mol * 38.00 ֯C = 30904.64 J

W = P * ΔV = 1atm * (948 - 876) L =  122 atmL

1atmL = 101.325 J

W = 122 * 101.325 J = 12361.65 J

 

ΔE = q + W = (30904.64 + 12361.65) J = 43266.29 J

Answer is 43266.29 J