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Question

A gas occupies $$500\ cm^{3}$$ at normal temperature. At what temperature will the volume of the gas be reduced by $$20\%$$ of its original volume, pressure being constant?


Solution

V = $$ 500 cm^3$$

Normal temperature, t = $$ 0^{\circ}C$$ = 0 + 273 K

$$ V_{1}$$ = Reduced volume + 20% of $$ 500 cm^3$$

= $$ \dfrac{20 \times 500}{100} = 100 cm^3$$

Net, $$ V_{1}$$ = 500 - 100 = $$ 400 cm^3$$

$$ T_{1}$$ = ?
P = $$ P_{1}$$

From gas equation,
$$ \dfrac{PV}{T} = \dfrac{P_1V_1}{T_1}$$

$$ \dfrac{P \times 500}{273} = \dfrac{P \times 400}{T_1}$$

$$ T_1 = \dfrac{273 \times 4}{5}$$
        = $$218.4$$
        = $$218.4 - 273$$
        = $$ 54.60^{\circ}C$$

Chemistry

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