The increase in internal energy = Energy supplied by heating + Work done on the system
ΔU = q + W
When a gas expands, work done W is negative
When a gas is compressed, work done W is positive
Positive or negative work done depends on whether the gas is compressed or expanded
Work is only done when the volume of a gas changes
The volume occupied by 1.00 mol of a liquid at 50 oC is 2.4 × 10-5 m3. When the liquid is vaporised at an atmospheric pressure of 1.03 × 105 Pa, the vapour has a volume of 5.9 × 10-2 m3.The latent heat to vaporise 1.00 mol of this liquid at 50 oC at atmospheric pressure is 3.48 × 104 J.Determine for this change of state the increase in internal energy ΔU of the system.
Step 1: Write down the first law of thermodynamics
ΔU = q + W
Step 2: Write the value of heating q of the system
This is the latent heat, the heat required to vaporise the liquid = 3.48 × 104 J
Step 3: Calculate the work done W
W = pΔV
ΔV = final volume − initial volume = 5.9 × 10-2 − 2.4 × 10-5 = 0.058976 m3
p = atmospheric pressure = 1.03 × 105 Pa
W = (1.03 × 105) × 0.058976 = 6074.528 = 6.07 × 103 J
Since the gas is expanding, this work done is negative
W = −6.07 × 103 J
Step 4: Substitute the values into first law of thermodynamics
ΔU = 3.48 × 104 + (−6.07 × 103) = 28 730 = 29 000 J (2 s.f.)
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