Question

In a certain region of space, there are only 5 molecules per cm^3 on an average and the temperature is 3 K. What is the average pressure of this very dilute gas ?

Answer 1

Question : In a certain region of space, there are only 5 molecules per $cm^3$ on an average and the temperature is 3 K. What is the average pressure of this very dilute gas ?

Answer :

The average pressure of this very dilute gas is $2.07\times 10^-16$ Pa.

Step-by-step explanation :

Here,

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

$V=5\:molecules/cm^3$

$T=3\:K$

$P=?$

No of molecules in 1 $cm^3$ = 5

No of molecules in 1 $m^3$ = $5\times 10^6$

$\because PV=nRT$

$\therefore P=\frac{n}{V}RT$ -----(1)

Now, the number of moles in pet unit volume,

⇒ $\frac{n}{V}=\frac{5\times 10^6}{6.022\times 10^23}$

Now, from equation (1),

⇒ $P=\frac{5\times 10^6}{6.02\times 10^23}\times 8.31\times 3$

⇒ $P=2.07\times 10^-16$ Pa

Answer 2

$\huge\red{Answer}$

n = 5/6.023×10²³

V = 1 cm³

= 10^-3 l

T = 3k

P×10^-3 = 5×0.0821×3/6.023×10²³

P-2.04×10^-21 atm

Average pressure of the gas = 2.04 x 10^-21 atm.

HOPE IT HELPS YOU !!