Medium
The density of water is $1000 \; kg \; m^{-3}$. The density of water vapour at $100^{\circ} C$ and $1 \; atm$ pressure is $0.6 \; kg \; m^{-3}$. The volume of a molecule multiplied by the total number gives what is called molecular volume. Estimate the volume of a water molecule.
(N/A) In the liquid phase,the molecules of water are quite closely packed. The density of a water molecule may therefore be regarded as roughly equal to the density of bulk water $= 1000 \; kg \; m^{-3}$.
To estimate the volume of a water molecule,we need to know the mass of a single water molecule.
We know that $1 \; mole$ of water has a mass approximately equal to $(2 + 16) \; g = 18 \; g = 0.018 \; kg$.
Since $1 \; mole$ contains about $6 \times 10^{23}$ molecules (Avogadro's number),the mass of a single molecule of water is $(0.018) / (6 \times 10^{23}) \; kg = 3 \times 10^{-26} \; kg$.
Therefore,a rough estimate of the volume of a water molecule is:
Volume of a water molecule $= (3 \times 10^{-26} \; kg) / (1000 \; kg \; m^{-3}) = 3 \times 10^{-29} \; m^3$.
Assuming the molecule is spherical,$V = (4/3) \pi r^3$,which gives a radius $r \approx 2 \times 10^{-10} \; m = 2 \; \mathring{A}$.
To estimate the volume of a water molecule,we need to know the mass of a single water molecule.
We know that $1 \; mole$ of water has a mass approximately equal to $(2 + 16) \; g = 18 \; g = 0.018 \; kg$.
Since $1 \; mole$ contains about $6 \times 10^{23}$ molecules (Avogadro's number),the mass of a single molecule of water is $(0.018) / (6 \times 10^{23}) \; kg = 3 \times 10^{-26} \; kg$.
Therefore,a rough estimate of the volume of a water molecule is:
Volume of a water molecule $= (3 \times 10^{-26} \; kg) / (1000 \; kg \; m^{-3}) = 3 \times 10^{-29} \; m^3$.
Assuming the molecule is spherical,$V = (4/3) \pi r^3$,which gives a radius $r \approx 2 \times 10^{-10} \; m = 2 \; \mathring{A}$.
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Similar Questions
Match List-$I$ with List-$II$:
List-$I$ List-$II$ $(A)$ Young's Modulus $(I)$ $[ML^{-1}T^{-1}]$ $(B)$ Torque $(II)$ $[ML^{-1}T^{-2}]$ $(C)$ Coefficient of Viscosity $(III)$ $[M^{-1}L^3T^{-2}]$ $(D)$ Gravitational Constant $(IV)$ $[ML^2T^{-2}]$
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $(A)$ Young's Modulus | $(I)$ $[ML^{-1}T^{-1}]$ |
| $(B)$ Torque | $(II)$ $[ML^{-1}T^{-2}]$ |
| $(C)$ Coefficient of Viscosity | $(III)$ $[M^{-1}L^3T^{-2}]$ |
| $(D)$ Gravitational Constant | $(IV)$ $[ML^2T^{-2}]$ |
Choose the correct answer from the options given below:
Choose the correct match:
List-$I$ List-$II$ $(i)$ Curie $(A)$ $MLT^{-2}$ $(ii)$ Light year $(B)$ $M$ $(iii)$ Dielectric strength $(C)$ Dimensionless $(iv)$ Atomic weight $(D)$ $T$ $(v)$ Decibel $(E)$ $ML^2T^{-2}$ $(F)$ $MT^{-3}$ $(G)$ $T^{-1}$ $(H)$ $L$ $(I)$ $MLT^{-3}I^{-1}$ $(J)$ $LT^{-1}$
| List-$I$ | List-$II$ |
|---|---|
| $(i)$ Curie | $(A)$ $MLT^{-2}$ |
| $(ii)$ Light year | $(B)$ $M$ |
| $(iii)$ Dielectric strength | $(C)$ Dimensionless |
| $(iv)$ Atomic weight | $(D)$ $T$ |
| $(v)$ Decibel | $(E)$ $ML^2T^{-2}$ |
| $(F)$ $MT^{-3}$ | |
| $(G)$ $T^{-1}$ | |
| $(H)$ $L$ | |
| $(I)$ $MLT^{-3}I^{-1}$ | |
| $(J)$ $LT^{-1}$ |
MediumIIT 1992
View SolutionThe mechanical equivalent of heat $J$ is
Easy
View SolutionSome physical constants are given in List-$I$ and their dimensional formulae are given in List-$II$. Match the following:
List-$I$ List-$II$ $(1)$ Planck's constant $(i)$ $[ML^{-1} T^{-2}]$ $(2)$ Gravitational constant (ii) $[ML^{-1} T^{-1}]$ $(3)$ Bulk modulus (iii) $[ML^2 T^{-1}]$ $(4)$ Coefficient of viscosity (iv) $[M^{-1} L^3 T^{-2}]$
The correct answer is:
| List-$I$ | List-$II$ |
| $(1)$ Planck's constant | $(i)$ $[ML^{-1} T^{-2}]$ |
| $(2)$ Gravitational constant | (ii) $[ML^{-1} T^{-1}]$ |
| $(3)$ Bulk modulus | (iii) $[ML^2 T^{-1}]$ |
| $(4)$ Coefficient of viscosity | (iv) $[M^{-1} L^3 T^{-2}]$ |
The correct answer is:
It is estimated that each $cm^2$ of Earth receives about $2 \text{ calorie}$ of heat energy per minute from the Sun. This is called the solar constant. The value of the solar constant in $S.I.$ units is:
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