Physics, asked by sarahshaikh6851, 11 months ago

A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m3. Find the pressure of the mixture.

Answers

Answered by jitendra420156
1

Therefore total pressure = 2,250 pa

Explanation:

Atomic mass: Atomic mass of a chemical component is the mass of chemical  component of 1 mole.

Atomic mass of O₂ is = 32 gram / mole

It means the weight of 1 mole is 32 gram.

Therefore the number of mole in 1.6 gram of O₂ is =\frac{1.6}{32} = 0.05 gram

Atomic mass of N₂ is 28 gram.

  It means the weight of 1 mole is 28 gram.

Therefore the number of mole in 1.6 gram of N₂ is \frac{2.80}{28} gram = 0.1 gram.

A ideal gas always flow  P V=nRT

\Rightarrow P=\frac{nRT}{v}

P = pressure

V = volume= 0.166 m³

n = no. of molecules

R= gas constant= 8.3 J/mole

T = temperature = 300 k

The partial pressure of O₂ is P_{O_2}=\frac{n_{O_2}RT}{v}

                                                       =\frac{0.05\times 8.3 \times 300}{0.166}

                                                      =750 pa

The partial pressure of N₂ is P_{N_2}=\frac{n_{N_2}RT}{v}

                                                       =\frac{0.1\times 8.3 \times 300}{0.166}

                                                      =1550 pa

Therefore total pressure = (750+1550)pa=2,250 pa

Answered by rahul123437
1

The Pressure of the oxygen and nitrogen mixture is 2250 Pa

Explanation:

Given data in the question  

V = 0.166 m^3  

T = 300 K

Mass of Oxygen (O_2 ) = 1.60 g

Molecular weight of Oxygen = 32 g

\mathrm{n}_{O}=\frac{1.60}{32}=0.05

Mass of Nitrogen (N_2) = 2.80 g

Molecular weight = 28 g

\mathrm{n}_{\mathrm{W}}=\frac{2.80}{28}=0.1

Partial O_2  pressure is given by

P_{0}=\frac{n_{O} R T}{V}=\frac{0.05 \times 8.3 \times 300}{0.166}=750

Partial N_2 pressure is given by

P_{N}=\frac{n_{n} R T}{V}=\frac{0.1 \times 8.3 \times 300}{0.166}=1500

Maximum pressure is the number of partial loads.

P=P_{N}+P_{0}=750+1500=2250 \mathrm{Pa}

The pressure of the mixture which contains 1.60 g of Oxygen and 2.80 g of Nitrogen with 300 K temperature is 2250 Pa in a vessel of volume 0.166 m^3

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