(A) The number of moles $\mu$ is given by the mass of water divided by its molar mass.$\mu = \frac{4.5 \,kg}{18 \times 10^{-3} \,kg/mol} = 250 \,mol$.

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(A) The number of moles $\mu$ is given by the mass of water divided by its molar mass.$\mu = \frac{4.5 \,kg}{18 \times 10^{-3} \,kg/mol} = 250 \,mol$.At standard temperature and pressure $(STP)$,$T = 273 \,K$ and $P = 10^5 \,Pa$.Using the ideal gas equation $PV = \mu RT$:$V = \frac{\mu RT}{P} = \frac{250 \times 8.314 \times 273}{10^5} \approx 5.67 \,m^3$.Rounding to the nearest provided option,the correct value is $5.6 \,m^3$.

Class 11 Physics Kinetic Theory of Gases Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

Medium

If the intermolecular forces vanish away,the volume occupied by the molecules contained in $4.5 \,kg$ of water at standard temperature and pressure will be

A
$5.6 \,m^3$
B
$4.5 \,m^3$
C
$11.2 \,litre$
D
$11.2 \,m^3$

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Kinetic Theory of Gases

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Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

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If the intermolecular forces vanish away,the volume occupied by the molecules contained in $4.5 \,kg$ of water at standard temperature and pressure will be

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