Question

If the pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1Kelvin it's final temperature must be?? I WANT FULL EXLANATION PLZ

Answer 1

Answer:

564 degrees

Explanation:

Answer 2

Answer:

The initial temperature was 248 K.

Explanation:

The container is a closed vessel, so its volume is constant.

That means we can apply Gay-Lussac's law.

According to which,

P₁/P₂  =   T₁/T₂ ,                                             ...(1)

where P₁ and P₂ are the initial and final pressures and T₁ and T₂ are the initial and final temperatures in Kelvin respectively.

Given that Pressure is increased by 4% (0.004)

So, P₂ = Initial Pressure + Increased pressure = (P₁ + 0.004P₁ = 1.004P₁)

T₁ in Celcius = (T₁ + 273)°C

Also,

T₂ is increased by 1°C.

So, T₂ = (T₁ + 273 + 1)°C    = (T₁ + 274)°C  

Putting these results in equation (1), we get:

P₁/1.004P₁ = (T₁ + 273)/(T₁ + 274)

⇒ 1/1.004 = (T₁ + 273)/(T₁ + 274)

⇒ (T₁ + 274) = 1.004(T₁ + 273) = 1.004T₁ + 274.1

⇒ 0.004T₁ = -0.1

⇒ T₁ = -0.1/0.004 = -25°C

Temperature in Kelvin = (273-25)K = 248 k