Question

Calculate the volume occupied by 5.25g of nitrogen at 26°c and 74.2cm of pressure

Answer 1

P= 74.2/76 atm

V=?

m=5.25 g

M = 28 g

T=273+26

T=299 K

PV=nRT

74.2/76 x V= m/M x  0.0821 x 299

V=5.25 x 0.0821 x 299 x 76 /74.2

V=4.71074201 L

Answer 2

Answer : The volume occupied will be, 4.74 L

Solution :

Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT\\\\V=\frac{wRT}{PM}$

where,

n = number of moles of gas

w = mass of nitrogen = 5.25 g

P = pressure of the gas = $74.2cmHg=\frac{74.2}{76}=0.97atm$

conversion : (1 atm = 76 cmHg)

T = temperature of the gas = $26^oC=273+26=299K$

M = molar mass of nitrogen gas = 28 g/mole

R = gas constant = 0.0821 Latm/moleK

Now put all the given values in the above equation, we get the volume.

$V=\frac{wRT}{PM}$

$V=\frac{5.25g\times 0.0821Latm/moleK\times 299K}{0.97atm\times 28g/mole}=4.74L$

Therefore, the volume occupied will be, 4.74 L