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Work Done in Isothermal Reversible Process

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Question

Calculate the magnitude of maximum work done (in $J$ ) in expanding 16 g of oxygen at 300 K and occupying a volume of $5 d m^{3}$ isothermally untill the volume becomes $25 d m^{3}$ .
(Given $l o g 5 = 0.6989$ )

2159 J

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2007 J

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1055 J

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2536 J

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Solution

The correct option is B 2007 J
Reversible work is maximum work.
$∴ W = - 2.303 n R T {log}_{10} (\frac{V_{2}}{V_{1}})$
where $V_{2}$ and $V_{1}$ are the final and initial volumes
n = moles of the gas
$W = 2.303 \times \frac{16}{32} \times 8.314 \times 300 log \frac{25}{5}$
$W = 2007.49 J$

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Work Done in Isothermal Reversible Process

Standard XII Chemistry

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Calculate the magnitude of maximum work done (in J) in expanding 16 g of oxygen at 300 K and occupying a volume of 5 dm3 isothermally untill the volume becomes 25 dm3. (Given log5=0.6989)

The correct option is B 2007 JReversible work is maximum work. ∴W=−2.303nRT log10(V2V1) where V2 and V1 are the final and initial volumes n = moles of the gas W=2.303×1632×8.314×300log255 W=2007.49 J

Calculate the magnitude of maximum work done (in $J$ ) in expanding 16 g of oxygen at 300 K and occupying a volume of $5 d m^{3}$ isothermally untill the volume becomes $25 d m^{3}$ .
(Given $l o g 5 = 0.6989$ )

The correct option is B 2007 J
Reversible work is maximum work.
$∴ W = - 2.303 n R T {log}_{10} (\frac{V_{2}}{V_{1}})$
where $V_{2}$ and $V_{1}$ are the final and initial volumes
n = moles of the gas
$W = 2.303 \times \frac{16}{32} \times 8.314 \times 300 log \frac{25}{5}$
$W = 2007.49 J$