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

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Solution

V = $500 c m^{3}$

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

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

= $\frac{20 \times 500}{100} = 100 c m^{3}$

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

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

From gas equation,
$\frac{P V}{T} = \frac{P_{1} V_{1}}{T_{1}}$

$\frac{P \times 500}{273} = \frac{P \times 400}{T_{1}}$

$T_{1} = \frac{273 \times 4}{5}$
= $218.4$
= $218.4 - 273$
= ${54.60}^{\circ} C$

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A gas occupies 500 cm3 at normal temperature. At what temperature will the volume of the gas be reduced by 20% of its original volume, pressure being constant?

V = 500cm3Normal temperature, t = 0∘C = 0 + 273 KV1 = Reduced volume + 20% of 500cm3= 20×500100=100cm3Net, V1 = 500 - 100 = 400cm3T1 = ?P = P1From gas equation,PVT=P1V1T1P×500273=P×400T1T1=273×45 = 218.4 = 218.4−273 = 54.60∘C

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