Question

calculate total pressure in a mixture of 8 gram Oxygen and 4 gram hydrogen confined in a vessel of of one 1 dm^3 at 27°C. R= 0.083​

Answer 1

Answer: 54atm or 54bar

Explanation:

Answer 2

The total pressure of the mixture is 55.42 atm

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For oxygen:

Given mass of oxygen = 8 g

Molar mass of oxygen = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen}=\frac{8g}{32g/mol}=0.25mol$

• For hydrogen:

Given mass of hydrogen = 4 g

Molar mass of hydrogen = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen}=\frac{4g}{2g/mol}=2mol$

To calculate the total pressure, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = ?

V = Volume of the gas = $1dm^3=1L$

T = Temperature of the gas = $27^oC=[27+273]K=300K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = Total number of moles = [0.25 + 2] = 2.25 moles

Putting values in above equation, we get:

$P\times 1L=2.25mol\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 300K\\\\P=\frac{2.25\times 0.0821\times 300}{1}=55.42atm$

Learn more about number of moles and ideal gases:

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