In a certain region of space there are only 5 molecules per `cm^(3)` on an average. The temperature there is 3 K. The pressure of this dilute gas is : `(k=1.38 times 10^(-23)J//K)`

Options

A. `20.7 times 10^(-17)N//m^(2)`
B. `15.3 times 10^(-15)N//m^(2)`
C. `2.3 times 10^(-10)N//m^(2)`
D. `5.3 times 10^(-5)N//m^(2)`

Answer

A

Explanation

From ideal gas equation
`" "PV=nkT`
`" "P=n/VkT`
Here, `n/V=5//cm^(3)=5 times 10^(6)//m^(3)`
`thereforeP=(5 times 10^(6)//m^(3))=(1.38 times 10^(-23)J//K) times 3K`
`" "P=20.7 times 10^(-17)N//m^(2)`