Question

A steel tank filled with 62.7 l of nitrogen gas at 85.0 atm and 19c is heated to 330c while the volume remains constant. what is the final gas pressure in atmospheres

Answer 1

Given:

The volume of the nitrogen gas in the steel tank = 62.7 L

The initial pressure inside the tank = 85 atm

The initial temperature = 19 °C = 19 + 273 K = 292 K

The final temperature inside the tank after heating = 330 °C = 330 + 273 K = 603 K

To find:

The final gas pressure in atmospheres

Solution:

According to the Ideal Gas Law, we have

$\boxed{\bold{PV = nRT}}$

We know that if the volume of the gas remains constant, then on increasing the temperature, the pressure of the enclosed gas also increases. This is known as the Gay-Lussac's Law.

$\frac{P}{T}$ = constant ....... when the volume of the gas remains constant

So, we can rewrite the formula as,

$\boxed{\frac{P1}{T1} = \frac{P2}{T2} }$

where

P1 = initial pressure

T1 = initial temperature

P2 = final pressure

T2 = final temperature

Now, on substituting the given values in the formula, we get

$\frac{85}{292} = \frac{P2}{603}$

⇒ P2 = $\frac{85\:*\:603}{292}$

⇒ P2 = $\frac{51255}{292}$

⇒ P2 = 175.53 atm

Thus, the final gas pressure in atmosphere is 175.53 atm.

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Answer 2

The final gas pressure in atmospheres is 175.53 atm.

Explanation:

Ideal gas law is given by the equation,

PV = nRT

For the given condition, the equation becomes,

P₁/T₁ = P₂/T₂

Where,

P₁ = Initial pressure = 85.0 atm

T₁ = Initial temperature = 19°C = 292 K

P₂ = Final pressure = ?

T₂ = Final temperature = 330°C = 603 K

On substituting the values, we get,

85/292 = P₂/603

P₂ = (85 × 603)/292

P₂ = 51255/292

∴ P₂ = 175.5 atm