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Work Done in Isothermal Reversible Process

Which of the ...

Question

Which of the following options are correct regarding : $q$ , $w$ and $△ U$ for the reversible isothermal expansion of one mole of an ideal gas at $27^{o} C$ from a volume of $10 d m^{3}$ to a volume of $20 d m^{3}$ ?

△ U = 0

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q = - w = 1729 J

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q = w = 1600 J

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△ U = 545 J

q = 0

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Solution

The correct option is B $q = - w = 1729 J$
$T = 27^{o} C = 27 + 273 = 300 K$
Since the process is reversible isothermal,
$△ U = 0$

Work done is given as:
$w = - 2.303 n R T l o g (\frac{V_{2}}{V_{1}})$

Putting all the values we get:

$w = - (2.303 \times 1 \times 8.314 \times 300) l o g (20 / 10) = - 1729 J$

From the first law, $△ U = q + w$
Since, $△ U = 0, q = - w = 1729 J$

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Work Done in Isothermal Reversible Process

Standard XI Chemistry

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Which of the following options are correct regarding : q, w and △U for the reversible isothermal expansion of one mole of an ideal gas at 27oC from a volume of 10 dm3 to a volume of 20 dm3?

The correct option is B q=−w=1729JT=27oC=27+273=300 K Since the process is reversible isothermal, △U=0 Work done is given as: w=−2.303 nRT log(V2V1) Putting all the values we get: w=−(2.303×1×8.314×300) log (20/10)=−1729 J From the first law, △U=q+w Since, △U=0,q=−w=1729 J

Which of the following options are correct regarding : $q$ , $w$ and $△ U$ for the reversible isothermal expansion of one mole of an ideal gas at $27^{o} C$ from a volume of $10 d m^{3}$ to a volume of $20 d m^{3}$ ?