Question

A gas at 283K has its temperature raised so that its volume is doubled the pressure remaining constant. What is its final temperature.​

Answer 1

Given :-

Initial temperature of gas = 10°C

volume is doubled the pressure remaining constant

To Find :-

Final temperature of the gas.

Solution :-

using Charles law at constant pressure .i.e.,

At constant pressure, the volume of the given mass of a gas is directly proportional to it's temperature on Kelvin scale.

In order words,

• V ∝ T

➠ V₁/V₂ = T₁/T₂

here, V₁ denotes Initial volume,V₂ denotes Final volume, T₁ denotes Initial temperature and T₂ Denotes Final temperature.

According to Question,

• T₁ = 283K
• V₂ = 2V₁

so,

➠ V₁/2V₁ = 283/T₂

➠ 2 = 283/T₂

➠ T₂ = 283 × 2

➠ T₂ = 566K

thus, the final temperature of the gas is 566K.

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#sanvi….

Answer 2

Solution :-

As per the given data ,

• V ' = 2 V
• Initial temperature (T) = 283 K
• Pressure remains constant

As per the ideal gas equation ,

➠ PV = nRT

On rearranging ,

→ V / T = nR /T

→ V / T = constant

Hence ,

→ V ∝ T

As per the Charles law ,

The volume of the ideal gas is directly proportional to the temperature when pressure is constant

Hence ,

→ V / V ' = T / T'

→ T ' = T x V' / V

→ T ' = 283 x 2V / V

→ T ' = 283 X 2

→ T ' = 566 K

∴ The final temperature of the mixture is 566 K