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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