Question

Calculate the number of molecules present in 350 cm³ of NH3 gas at 273K and 2 atmospheric pressure.

Answer 1

According to ideal gas equation, PV = n RT

The given data are V = 350 cm3 or 350x10^-6 m3;

T = 273 K;

R= 8.31441 J K-1 mol-1;

P = 2 atm ;
But 1 atmosphere = 101325 Pa
hence 2 atmosphere = 202650 Pa

n = PV/RT

= (202650 Pa x 350x10^-6 m3)/ 8.314 J K-1 mol-1 x 298 K

= (70927500 x10^-6)/2477.572

n = 28627.8 x10^-6 mol
or 0.03 mol