Question

Estimate the total number of air molecules(inclusive of oxygen, nitrogen, water and other constituents) in a room of capacity 25 m^3 at a temperature of 27° C and 1 atm pressure.

Answer 1

$\mathbb{\bold{\red{ANSWER}}}$

QUESTION==>>>>>>>>>>>

Estimate the total number of air molecules(inclusive of oxygen, nitrogen, water and other constituents) in a room of capacity 25 m^3 at a temperature of 27° C and 1 atm pressure.

Given=>..

According to the question,, a'n' moles of gas present in room of temperature 27℃ and of volume 25$25m^{3}$,, need to find out the no of moles of gas..

so,,

Volume = $25m^{3}$

temperature= 27℃=> K

°K = 273 + 27= 300K...

Pressure = 1 atm= 1.01 × $10^{5}$

To find,, n=?..

R = 8.314$\frac{J}{Kmol}$

>>>>

we have,,

PV = n R T

n = $\frac{PV}{RT}$

n = $\frac{1.01\times10^{5}\times25\times6.022\times10^{23}}{8.313\times300}$

n = $\frac{152.055\times10^{23+5}}{2493.9}$

n = 6.09$10^{26}$moles

Answer 2

Question : Estimate the total number of air molecules(inclusive of oxygen, nitrogen, water and other constituents) in a room of capacity 25 $m^3$ at a temperature of 27° C and 1 atm pressure.

Answer :

The total number of air molecules(inclusive of oxygen, nitrogen, water and other constituents) in a room of capacity 25 $m^3$ at a temperature of 27° C and 1 atm pressure is $6.1\times 10^26$ .

Step-by-step explanation :

Given,

$R=8.31\:J/mol\:K$

$V=25\:m^3$

$T=300\:K$

$P=1.01\times 10^5\:N/m^2$

Using the formula,

⇒ $PV=nRT$

⇒ $n=\frac{PV}{RT}$

⇒ $n=\frac{1.01\times 10^5\times 25}{8.31\times 300}$

⇒ $n=\frac{1.01\times 10^5}{8.31\times 12}$

⇒ $n=\frac{1.01\times 10^5}{99.72}$

Therefore, the number of air molecules,

⇒ $\frac{1.01\times 10^5}{99.72}\times 6.022\times 10^23$

⇒ $6.1\times 10^26$