Question

Calculate volume occupied by 16 g o2 at 300 k and 8.31 mpa​

Answer 1

Answer:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number 6.023\times 10^{23}6.023×1023 of particles.

According to the ideal gas equation:

PV=nRTPV=nRT

P = Pressure of the gas = 8.31 mPa = 8.20\times 10^{-8}8.20×10−8 atm  (1mPa=9.87\times 10^{-9}atm(1mPa=9.87×10−9atm

V= Volume of the gas = ?

T= Temperature of the gas = 300 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas=\frac{\text {given mass}}{\text {molar mass}}=\frac{16g}{32g/mol}=0.5molesmolar massgiven mass=32g/mol16g=0.5moles

V=\frac{nRT}{P}=\frac{0.5\times 0.0821\times 300}{8.20\times 10^{-8}}=1.50\times 10^{8}LV=PnRT=8.20×10−80.5×0.0821×300=1.50×108L