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Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

(e) the number of air molecules in your classroom.


Let, the volume of the classroom be V.

One mole of air at NTP occupies 22.4 l.
i.e., 22.4 × 10
–3 m3volume.

Number of molecules in one mole = 6.023 × 1023 

∴ Number of molecules in room of volume V,

= 6.023 × 1023 × V / 22.4 × 10-3 

=  134.915 × 10
26 V 

=  1.35 × 10
28 V 

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The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data :
Mass of the Sun = 2.0 ×10 30kg,
Radius of the Sun = 7.0 × 108m


Mass space of space the space Sun comma space straight M space equals space 2 cross times 10 to the power of 30 space kg space

Raidus comma space straight r equals space 7 cross times 10 to the power of 8 space straight m space

Therefore comma space

Density space of space the space Sun space equals space straight M over straight V space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator straight M over denominator begin display style bevelled 4 over 3 end style πr cubed end fraction space equals space fraction numerator 3 cross times 2 cross times 10 to the power of 30 over denominator 4 straight pi space left parenthesis 7 cross times 10 to the power of 8 space straight m space right parenthesis cubed end fraction space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1400 space kg divided by straight m to the power of 3 space end exponent

That space is comma space

Mass space density space of space Sun space is space in space the space range space of space mass space densities
of space solids divided by liquids space and space not space gases. space
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When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72” of arc. Calculate the diameter of Jupiter.

Given comma space

Angular space diameter comma space straight theta space equals space 35.72 " space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 35.72 " over denominator 3600 end fraction cross times straight pi over 180 space rad space

Distance space of space Jupiter space from space Earth comma space straight d space equals space 824.7 space cross times space 10 to the power of 9 space straight m space

So comma space diameter space of space Jupiter comma space straight D space equals space straight theta space cross times straight d space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.73 space cross times space 10 to the power of negative 4 end exponent space cross times space 824.7 space cross times 10 to the power of 9 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.43 space cross times 10 to the power of 8 space straight m
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Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase : 970 kg m–3. Are the two densities of the same order of magnitude ? If so, why?

Diameter space of space sodium space atom space is space equal space to space the space size space of space sodium space atom.

That space is space straight D space equals space 2.5 space straight Å space

So comma space radius space of space sodium space atom comma space straight r space equals space left parenthesis 1 divided by 2 right parenthesis space cross times space 2.5 space straight Å space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.25 space straight Å space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.25 space cross times space 10 to the power of negative 10 end exponent space straight m space

Volume space of space sodium space atom comma space straight V space equals space 4 over 3 πr cubed space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 over 3 cross times 3.14 cross times left parenthesis 1.25 cross times 10 to the power of negative 10 end exponent right parenthesis cubed space
According space to space the space Avogadro space hypothesis comma
one space mole space of space sodium space contains space 6.023 space cross times space 10 to the power of 23 space atoms space
and space has space straight a space mass space of space 23 space straight g space or space 23 space cross times space 10 to the power of – 3 end exponent space kg.

therefore space space Mass space of space one space atom space equals space fraction numerator space 23 space cross times space 10 to the power of – 3 end exponent space over denominator space 6.023 space cross times space 10 to the power of 23 end fraction space equals space straight m subscript 1 space

Density space of space sodium space atom comma space straight rho space equals space straight m subscript 1 over straight V space

Putting space values space in space this space equation comma space

Density space of space sodium space atom comma space straight rho space equals 4.67 space cross times space 10 to the power of negative 3 end exponent space Kg space straight m to the power of negative 3 end exponent space

Density space of space sodium space in space crystalline space phase space equals space 970 space kg space straight m to the power of negative 3 end exponent space
Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. This is because in solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase. 
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It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s.

Total time for which the two watches run, T = 100 years

Diffrence in the time = 0.02 sec 

Error in 1 sec = fraction numerator 0.02 over denominator 3.1536 cross times 10 to the power of 9 end fraction space = 6.34 space cross times 10 to the power of negative 12 end exponent space straight s space almost equal to 10 to the power of negative 11 end exponent straight s

Thus, accuracy of 1 part is 1011.

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