Question

calculate the density of O2 at 25 degree Celsius and 1.05 ATM pressure

Answer 1

As we know pv=nrt

Pv=wt\molec.w *rt

When volume goes on right side it becomes
P=density divide by molec weight * rt
Density = p * molecular wt divide by r(0.082)
temp (273+25)

d = 1.05* 32 (divide by) 0.082*298

DENSITY = 1.375

Answer 2

Given :

Temperature (T) = 25°C = (273 + 25) K = 298 K

Pressure (P) = 1.05 atm

To Find :

Density of $O_{2}$

Solution :

We know, PV = nRT ( n = no. of moles and R = molar gas constant)

      No. of moles (n) = $\frac{Given Mass (w)}{Molar Mass (M)}$

So,   PV = $\frac{w}{M}$×R×T

 or,      P = $\frac{w}{MV}$ ×R×T

 or,      P = $\frac{Density (D)}{Molecular Mass (M)} RT$  (as density = $\frac{mass}{volume}$)

 or,      D =  $\frac{PM}{RT}$

 or,      D = $\frac{1.05 * 32}{0.082 * 298}$

 or,      D = $\frac{33.6}{24.436}$

∴        D = 1.375 gm/L

∴ The density of O$_{2}$ at 25°C and 1.05 atm pressure is 1.375 gm/L