120 cm-3 of a gas is taken at 27-3 K- The temperature is then raised to 0- What is the new volume of the gas at Isobaric condition-1300 cm-3-1400 cm-3-1200 cm-3-1100 cm-3-

The correct option is C 1200 cm^3 By Charle's Law V_1/T_1 = V_2/T_2 V_1 = 120 cm^3 T_1 = 27.3 K T_2 = 273 + 0 = 273 K 120/27.3 = V_2/273 ∴ V_2 = 1200 cm^3 ...

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Standard IX

Chemistry

Temperature of a Gas

120 cm^3 of a...

Question

$120 c m^{3}$ of a gas is taken at 27.3 K. The temperature is then raised to $0^{\circ}$ . What is the new volume of the gas at Isobaric condition?

$1100 c m^{3}$

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$1200 c m^{3}$

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$1300 c m^{3}$

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$1400 c m^{3}$

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Solution

The correct option is B
$1200 c m^{3}$

By Charle's Law -

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$V_{1} = 120 c m^{3}$

$T_{1} = 27.3 K$

$T_{2} = 273 + 0 = 273 K$

$\frac{120}{27.3} = \frac{V_{2}}{273}$

$∴ V_{2} = 1200 c m^{3}$

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120cm3 of a gas is taken at 27.3 K. The temperature is then raised to 0∘. What is the new volume of the gas at Isobaric condition?

The correct option is B 1200cm3 By Charle's Law - V1T1=V2T2 V1=120cm3 T1=27.3K T2=273+0=273K 12027.3=V2273 ∴ V2=1200cm3

$120 c m^{3}$ of a gas is taken at 27.3 K. The temperature is then raised to $0^{\circ}$ . What is the new volume of the gas at Isobaric condition?

The correct option is B
$1200 c m^{3}$

By Charle's Law -

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$V_{1} = 120 c m^{3}$

$T_{1} = 27.3 K$

$T_{2} = 273 + 0 = 273 K$

$\frac{120}{27.3} = \frac{V_{2}}{273}$

$∴ V_{2} = 1200 c m^{3}$