Sum

**Answer the following question.**

Calculate the maximum work when 24 g of O_{2} are expanded isothermally and reversibly from the pressure of 1.6 bar to 1 bar at 298 K.

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#### Solution

**Given:**

Mass of O_{2} = 24 g

Initial pressure = P_{1} = 1.6 bar

Final pressure = P_{2} = 1 bar

Temperature = T = 298 K

**To find:** Maximum work (W_{max})

**Formula: **`"W"_"max" = - 2.303 "nRT" log_10 "P"_1/"P"_2`

**Calculation: **

Number of moles of O_{2} = n = `(24 "g")/(32 "g mol"^-1)` = 0.75 mol

Gas constant = R = 8.314 J K^{–1} mol^{–1}

Now, using formula,

`"W"_"max" = - 2.303 "nRT" log_10 "P"_1/"P"_2`

= - 2.303 × 0.75 mol × 8.314 J K^{-1}mol^{-1} × 298 K × `log_10 1.6/1`

= - 2.303 × 0.75 × 8.314 J × 298 × 0.2041

= - 873.4 J

The maximum work done is – 873.4 J.

Concept: Concept of Maximum Work

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