Question

In 2 L container, 10²³ molecules of NO₂ are filled at 27°C. Calculate the pressure in container?

Answer 1

Given, T = 27+ 273 = 300 K; V = 2.0 L 

R = 0.0821 L atm K^-1mol^-1 

Number of moles of NO₂ 

_{\( = \frac{10^{23}}{6.022\times 10^{23}}\)} 

0.166 mol 

From ideal gas equation,

PV = nRT

\(P = \frac{nRT}{V}\)

\( = \frac{0.166\ mol\times 0.0821\ L\ atm\ K^{-1}\ mol^{-1}\times 300\ K}{2.0\ L}\) 

= 0.204 atm 

So, pressure in container = 0.205 atm.