Gas law questions 2-5. Will give brainliest.

Explanation:

2)

Given data:

Initial volume = 1.50 L

Initial temperature = 20°C

Final temperature = 100°C +20°C = 120°C

Final volume = ?

Solution:

Initial temperature = 20°C = 20+273 = 293 K

Final temperature = 120+273 = 393 K

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1.50 L × 393 K / 293 K

V₂ = 589.5 L.K /293 K

V₂ = 2 L

3)

Given data:

Initial volume = 2.75 dm³

Initial pressure = 455 kpa (455/101 = 4.5 atm)

Initial temperature = 100 °C (100 +273 = 373 K)

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 4.5 atm × 2.75 dm³  × 273 K / 373 K × 1 atm

V₂ = 3378.375 atm. dm³.K / 373 K.atm

V₂ = 9 dm³

4)

Given data:

Total pressure of mixture = 1023 torr

Pressure of nitrogen = 544 torr

Pressure of oxygen = ?

Solution:

According to the Dalton law of partial pressure,

The pressure of mixture of a gas is the equal to the sum of partial pressure of individual gases.

P(total) = P₁+ P₂ + P₃ ++Pₙ

In given question total pressure is 1023 torr and partial pressure of nitrogen is 544 torr.

P(total) = P(N₂)+ P(O₂)

1023 torr = 544 torr +  P(O₂)

P(O₂) = 1023 torr - 544 torr

P(O₂) = 479 torr

5)

Given data:

Number of moles of gas = 2.43 mol

Temperature of gas = 35 °C (35+273 = 308 K)

Pressure of gas = 789 torr ( 789/760 = 1.03 atm)

Volume of gas = ?

Solution:

PV = nRT

R = general gas constant = 0.0821 atm.L/ mol.K

V = nRT/P

V = 2.43 mol ×0.0821 atm.L/ mol.K× 308 K / 1.03 atm

V = 61.45 L / 1.03

V = 59.66 L