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Entropy Change in Isothermal Process

Calculate q i...

Question

Calculate $q$ (in kilo joules), for the reversible isothermal expansion of two mole of an ideal gas at $27^{o} C$ from a volume of $10 d m^{3}$ to a volume of $30 d m^{3}$ .

0.89 k J

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3.4 k J

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1.7 k J

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5.4 k J

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Solution

The correct option is D $5.4 k J$
Work done for an isothermal reversible expansion is given as:
$w = - 2.303 n R T l o g \frac{V_{2}}{V_{1}}$

Given : $V_{1} = 10 L$
$V_{2} = 30 L$
$T = 27^{o} C$ = $27 + 273 = 300 K$
so,
$w = - 2 \times 8.314$ $\times 300 \times 2.303$ $l o g \frac{30}{10}$
$w = - 2 \times 8.314$ $\times 300 \times 2.303$ $l o g 3$
$w = - 5481.3 J$
Since the process is isothermal $△ U = 0$ ,
From the first law of thermodynamics,
$△ U = q + w$
$q = - w$
$q = 5481.3 J$
$q = 5.4 k J$

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Entropy Change in Isothermal Process

Standard XII Chemistry

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Calculate q (in kilo joules), for the reversible isothermal expansion of two mole of an ideal gas at 27oC from a volume of 10 dm3 to a volume of 30 dm3.

Calculate $q$ (in kilo joules), for the reversible isothermal expansion of two mole of an ideal gas at $27^{o} C$ from a volume of $10 d m^{3}$ to a volume of $30 d m^{3}$ .