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A system of gas at low density has an initial pressure of 1.90 × 10 5 1.90×105 Pa and occupies a volume of 0.18 m³. The slow addition of 965 J of heat to the system causes it to expand isobarically to a volume of 0.51 m³. What is the change in the internal energy of the system?

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Answer:

ΔU = -6.2 × 10⁴ J

Explanation:

The system absorbs 965 J of heat, that is, q = 965 J.

The work (w) can be calculated using the following expression.

w = -P . ΔV

where,

P is the external pressure

ΔV is the change in the volume

w = - (1.90 × 10⁵ N/m²) × (0.51 m³ - 0.18 m³) = -6.3 × 10⁴ J

The change in the internal energy (ΔU) is:

ΔU = q + w = 965 J + (-6.3 × 10⁴ J) = -6.2 × 10⁴ J

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