. An ideal gas expands from a volume of


of 6dm3 to 16dm3 against the constant external pressure of 2.026×105 Nm-2. Find enthalpy change if Δu is 418J.

 

Best Answer

• Given:

V1 = 6dm3 = 6 × (0.1m)3 = 6 × 10-3m3

V2 = 16dm3 = 16 × 10-3m3

Pext = 2.026 × 105Nm-2

∆u = 418J

• Work done is,

Work done = - P∆V

W = -2.026 × 105 (V2 – V1)

W = -2.026 × 105 (6 × 10-3 – 16 × 10-3) Nm

W = -2.026 × 105 (-10 × 10-3) joule

W = -2.026 × 103 joule

• Enthalpy change, which is denoted as ∆H is

∆H = ∆u + P∆H

∆H = 418 + 2.026 × 103 J

∆H = 2444J

 

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