Question

A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm. The pressure gauge of cylinder indicates 12atm at 27°C. Due to sudden fire in the building the temperature starts rising. The temperature at which the cylinder will explode is * 42.5°C 67.8°C 99.5°C 25.7°C​

Answer 1

Answer :-

The cylinder will explode at a temperature of 99.5°C .

Explanation :-

We have :-

→ Initial pressure (P₁) = 12 atm

→ Initial temperature (T₁) = 27°C

→ Final pressure (P₂) = 14.9 atm

______________________________

Firstly, let's convert the unit of initial temperature from °C to K .

⇒ 0°C = 273 K

⇒ 27°C = (273 + 27) K

⇒ 300 K

According to Gay Lussac's Law we know that at contsant volume, pressure of a given amount of gas is directly proportional to temperature in "Kelvin" . So, now we have :-

P₁/T₁ = P₂/T₂

⇒ 12/300 = 14.9/T₂

⇒ 0.04 = 14.9/T₂

⇒ 0.04T₂ = 14.9

⇒ T₂ = 14.9/0.04

⇒ T₂ = 372.5 K

______________________________

Finally, we will convert the unit of obtained temperature (T₂) from K to °C .

⇒ 273 K = 0°C

⇒ 372.5 K = (372.5 - 273) °C

⇒ 99.5°C

Answer 2

Given :-

A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm. The pressure gauge of the cylinder indicates 12atm at 27°C.

To Find :-

The temperature at which the cylinder will explode is

Solution :-

T₁ = 273 + T₁(C)

T₁ = 273 + 27

T₁ = 300 K

Now

P₁/P₂ = T₁/T₂

T₂ = P₁ × T₁/P₂

T = 14.9 × 300/12

T = 149 × 300/120

T = 149 × 30/12

T = 372.5 K

Now

C = K - 273

C = 372.5 - 273

C = 99.5° C

Hence

Option C is correct

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