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प्रश्न
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure
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उत्तर १
Volume of the room, V = 25.0 m3
Temperature of the room, T = 27°C = 300 K
Pressure in the room, P = 1 atm = 1 × 1.013 × 105 Pa
The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as:
PV = kBNT
Where,
KB is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1
N is the number of air molecules in the room
`:. N = (PV)/(k_BT)`
`= (1.013xx10^5xx25)/(1.38xx10^(-23)xx300) = 6.11 xx 10^(26)` molecules
Therefore, the total number of air molecules in the given room is 6.11 × 1026.
उत्तर २
Here, Volume of room, `V= 25.0 m^3`, temperature, `T = 27 ^@C = 300 K` and
Pressure, `P = 1 "atm" = 1.01 xx 10^5 Pa`
According to gas equation, `PV = muRT = muN_A.k_BT`
Hence, total number of air molecules in the volume of given gas
`N = mu.N_A = PV/K_BT`
`:. N = (1.01 xx 10^5 xx25.0)/((1.38 xx 10^(-23))xx300) = 6.1 xx 10^(26)`
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|
|
Column A |
Column B |
|
(a) |
cm3 |
(i) Pressure |
|
(b) |
Kelvin |
(ii) Temperature |
|
(c) |
Torr |
(iii) Volume |
|
(d) |
Boyle's law |
(iv) `"V"/"T" = ("V"_1)/("T"_1)` |
|
(a) |
Charles's law |
(v) `"PV"/"T" = ("P"_1 "V"_1)/"T"_1` |
|
|
|
(vi) PV = P1V1 |
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