Don't forget to try our free app -

Agile Log
,
which helps you track your time spent on various projects and tasks, :)

Try It Now
**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:

V_{1} = 0.5 lit

V_{2 }= 2.0 lit

P = 1 atmp

q = 29.0 calories

ΔV = V_{2} –V_{1} = 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

ΔS_{vap} = H / T

ΔS_{vap} = 10400 / (100 + 273) since ( 1 Kelvin = ° C + 273)

ΔS_{vap} = 10400 / 373

ΔS_{vap} = 27.88 cal / Kelvin