Current Location  >  Formulas in Chemistry > Thermodynamics > Thermodynamics (Examples )

Thermodynamics (Examples )

Example 1: Calculate the internal energy of a system, which gets 100 calories heat from the surrounding and work done by the system is 44 calories.

Here, q = 100 and w = – 44
So:
ΔE = q + w
ΔE = 100 – 44
ΔE = 56 calories

Example 2: Calculate the internal energy of a system, where the volume of a gas at atmospheric pressure was 0.5 lit. The gas obtained 29.0 calories heat from surrounding and its volume becomes 2.0 lit at 1 atmospheric pressure. (1 lit.atmp = 24.21 calories)

We have:
V1 = 0.5 lit
V2 = 2.0 lit
P = 1 atmp
q = 29.0 calories

ΔV = V2 –V1 = 2.0-0.5 = 1.5 lit

w = PΔV
w = 1 x 1.5
w = 1.5 lit. atmp
 
since 1 lit. atmp = 24.21 calories, we have:
1.5 lit. atmp = 24.21 x 1.5 = 36.32 calories

The work done by the system is therefore:
w = -36.32 calories
 
ΔE = q + Δw
ΔE = 29.0 – 36.32
ΔE = - 7.32 calories

Example 3: Calculate the heat capacity of ethanol, when the temperature of ethanol is increased by 2 °C by supply of 23.40 calorie.

Heat capacity C = Heat absorbed / Temperature difference = 23.40/2 = 11.70 Kcal/°C

Example 4: Calculate the change in entropy when two moles of water are converted into its vapor at that temperature by boiling at 100 °C? The molar heat of vaporization of water is 5200 cal/mole.

1 mole of water = 5200 cal require to vaporize
2 mole of water = 2 x 5200 = 10400 calorie

Here, the system is gaining energy, so we have:

ΔH = +10400 calories
ΔSvap = H / T
ΔSvap = 10400 / (100 + 273)  since ( 1 Kelvin = ° C + 273)
ΔSvap = 10400 / 373
ΔSvap = 27.88 cal / Kelvin

Web-Formulas.com © 2017 | Contact us | Terms of Use | Privacy Policy | Yellow Sparks Network
Web Formulas