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Byju's Answer

Standard X

Chemistry

Gay Lussac's Law

The pressure ...

Question

The pressure of air in the bulb of a constant volume gas thermometer at $0^{\circ}$ C and $100^{\circ}$ C are 73.00 cm and 100 cm of mercury respectively. Calculate the pressure at room temperature $20^{\circ}$ C.

78.4 c m

of Hg

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88.4 c m

of Hg

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98.4 c m

of Hg

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108.4 c m

of Hg

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Solution

The correct option is A $78.4 c m$ of Hg
$t = \frac{P_{t} - P_{0}}{P_{100} - P_{0}} \times 100^{\circ} C$
$%$ increase in pressure from $P_{0}$
$20^{\circ} C = \frac{P_{t} - 73}{100 - 73} \times 100$
$= 78.4 c m$ of Hg

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The pressure of air in the bulb of a constant volume gas thermometer at 0∘C and 100∘C are 73.00 cm and 100 cm of mercury respectively. Calculate the pressure at room temperature 20∘C.

The correct option is A 78.4 cm of Hgt=Pt−P0P100−P0×100∘ C % increase in pressure from P0 20∘ C=Pt−73100−73×100 =78.4 cm of Hg

The pressure of air in the bulb of a constant volume gas thermometer at $0^{\circ}$ C and $100^{\circ}$ C are 73.00 cm and 100 cm of mercury respectively. Calculate the pressure at room temperature $20^{\circ}$ C.

The correct option is A $78.4 c m$ of Hg
$t = \frac{P_{t} - P_{0}}{P_{100} - P_{0}} \times 100^{\circ} C$
$%$ increase in pressure from $P_{0}$
$20^{\circ} C = \frac{P_{t} - 73}{100 - 73} \times 100$
$= 78.4 c m$ of Hg