A gas is compressed at constant pressure 2.0 × 10⁵ Pa from 3.0 × 10⁻³ m³ to 2.0 × 10⁻³ m³ and loses 80 J by heating. State Q, W_by and W_on, then calculate ΔU using Cambridge's convention. Separately, an ideal gas is compressed isothermally with 120 J of work done on it; state Q.

Question

Accepted Answer

Q = −80 J; W_by = −200 J; W_on = +200 J; ΔU = +120 J. Isothermal case: Q = −120 J. ΔV = 2.0 × 10⁻³ − 3.0 × 10⁻³ = −1.0 × 10⁻³ m³. W_by = pΔV = −200 J, so W_on = +200 J. Energy leaves by heating, so Q = −80 J. ΔU = Q + W_on = −80 + 200 = +120 J. For the separate ideal-gas isothermal compression, ΔU = 0, so Q = −W_on = −120 J: 120 J leaves by heating.

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A gas is compressed at constant pressure 2.0 × 10⁵ Pa from 3.0 × 10⁻³ m³ to 2.0 × 10⁻³ m³ and loses 80 J by heating. State Q, W_by and W_on, then calculate ΔU using Cambridge's convention. Separately, an ideal gas is compressed isothermally with 120 J of work done on it; state Q.

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A gas is compressed at constant pressure 2.0 × 10⁵ Pa from 3.0 × 10⁻³ m³ to 2.0 × 10⁻³ m³ and loses 80 J by heating. State Q, Wby and Won, then calculate ΔU using Cambridge's convention. Separately, an ideal gas is compressed isothermally with 120 J of work done on it; state Q.

What a complete response should do

Your self-assessment

A gas is compressed at constant pressure 2.0 × 10⁵ Pa from 3.0 × 10⁻³ m³ to 2.0 × 10⁻³ m³ and loses 80 J by heating. State Q, W_by and W_on, then calculate ΔU using Cambridge's convention. Separately, an ideal gas is compressed isothermally with 120 J of work done on it; state Q.