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Work Done in Isothermal Reversible Process

Calculate q i...

Question

Calculate $q$ (in joules), for the reversible isothermal expansion of one mole of an ideal gas $27^{o} C$ from a volume of 10 $d m^{3}$ to a volume of $20 d m^{3}$ .

1729 J

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1504 J

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1432 J

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1830 J

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Solution

The correct option is A 1729 J
work is given as
$w = - 2.303 n R T l o g \frac{V_{2}}{V_{1}}$

Given : $V_{1} = 10 L$
$V_{2} = 20 L$
$T = 27^{o} C$ = $27 + 273 = 300 K$
so,
$w = - 1 \times 8.314$ $\times 300 \times 2.303$ $l o g \frac{20}{10}$
$w = - 1729 J$
since process is isothermal $△ U = 0$
from first law of thermodynamic,
$△ U = q + w$
$q = - w$
$q = 1729 J$

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Work Done in Isothermal Reversible Process

Standard XIII Chemistry

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Calculate q (in joules), for the reversible isothermal expansion of one mole of an ideal gas 27oC from a volume of 10 dm3 to a volume of 20 dm3.

The correct option is A 1729 Jwork is given as w=−2.303nRT logV2V1 Given : V1=10 L V2=20 L T=27oC = 27+273=300K so, w=−1×8.314 ×300×2.303 log2010 w=−1729 J since process is isothermal △U=0 from first law of thermodynamic, △U=q+w q=−w q=1729 J

Calculate $q$ (in joules), for the reversible isothermal expansion of one mole of an ideal gas $27^{o} C$ from a volume of 10 $d m^{3}$ to a volume of $20 d m^{3}$ .