Question

Calculate the total pressure in a mixture of 16 g of oxygen and 4g of Hydrogen confined in a vessel of 1dm-3 at 27 degree celsius. (Molar mass of oxygen 32 Hydrogen 2 R=0.083bar dm
3 K-1mol-1)

Answer 1

16 g  = 1/2 mole of O2                       4 g =  2 mole of H2
     V = volume = 1 d / m^3        T = 27 C = 300 K
 
 P V = n R T          =>  P1  V  +  P2 V  = n1 R T  + n2 R T
                                      P1 is due to n1 moles O2  and P2 is due to n2 moles of H2 
P1 + P2 =  (RT/V) (n1+n2) =    (0.083 dm3  * 300 / 1 dm^-3 ) 2.5
             = 62.25 bar

Answer 2

Answer : The total pressure of the mixture of gas is 62.25 bar.

Explanation :

First we have to determine the moles of oxygen and hydrogen gas.

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{4g}{2g/mole}=2moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{16g}{32g/mole}=0.5moles$

Now we have to calculate the total pressure of the mixture of gas.

Using ideal gas equation:

$PV=nRT$

As, the moles is an additive property. So,

$PV=(n_{H_2}+n_{O_2})RT$

where,

P = pressure of gas = ?

V = volume of gas = $1dm^3$

T = temperature of gas = $27^oC=273+27=300K$

$n_{H_2}$ = number of moles of hydrogen gas = 2

$n_{O_2}$ = number of moles of oxygen gas = 0.5

R = gas constant = $0.083bar.dm^3K^{-1}mol^{-1}$

Now put all the given values in the ideal gas equation, we get:

$P\times (1dm^3)=(2+0.5)mole\times (0.083bar.dm^3K^{-1}mol^{-1})\times (300K)$

$P=62.25bar$

Therefore, the total pressure of the mixture of gas is 62.25 bar.