Question

at 273k temperature and 1 bar pressure volume of gas increases 20% then what temperature required?​

Answer 1

Answer:-

Given:

Initial Temperature (T₁) = 273° K

Pressure = 1 bar

Let the Initial Volume of the gas be V.

The volume of the gas is increased by 20% .

⟹ Final Volume (V₂) = V * (100 + 20)/100

⟹ V₂ = 120V/100

We know that,

Charle's law of gases states that at constant pressure the Volume of a gas is directly proportional to the temperature.

⟹ V ∝ kT

⟹ V/T = k (constant).

⟹ V₁/T₁ = V₂/T₂

⟹ V/273 = (120V/100) / (T₂)

⟹ T₂ * V = (120V/100) * 273

⟹ T₂ = 120V/100 * 273 * 1/V

⟹ T₂ = 327.6° K

Therefore, the temperature of the gas increases to 327.6° K when volume of the gas increases by 20%.