Two moles of an ideal gas are compressed to half of their volume at a temperature of 300 K . The work done in the process is

-3457.6 J

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-5186.6 J

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3457.6 J

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5186.6 J

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Solution

The correct option is A -3457.6 J
Given data:
$T_{1} = T_{2} = 300 K and n = 2$

The work done is

$W = n R T_{1} l n \frac{V_{2}}{V_{1}}$

$= 2 \times 8.314 \times 300 l n \frac{1}{2}$

$= - 3457.6 J$

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Two moles of an ideal gas are compressed to half of their volume at a temperature of 300 K . The work done in the process is

The correct option is A -3457.6 JGiven data: T1=T2=300Kandn=2 The work done is W=nRT1lnV2V1 =2×8.314×300 ln12 =−3457.6J

The correct option is A -3457.6 J
Given data:
$T_{1} = T_{2} = 300 K and n = 2$

The work done is

$W = n R T_{1} l n \frac{V_{2}}{V_{1}}$

$= 2 \times 8.314 \times 300 l n \frac{1}{2}$

$= - 3457.6 J$