Question

Helium gas in a 2.00 L cylinder is under 1.03 atm pressure. At 37.1°C, that same gas sample has a pressure of 2.29 atm. What was the initial temperature of the gas in the cylinder?

Answer 1

Answer:

use ideal gas law

Explanation:

Answer 2

Given:

Helium gas in a 2.00 L cylinder is under 1.03 atm pressure

At 37.1°C, that same gas sample has a pressure of 2.29 atm

To find:

The initial temperature of the gas in the cylinder

Solution:

We have the equation for the Ideal Gas Law as:

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

where

P = Pressure

V = Volume

n = no. of moles

R = Ideal Gas constant

T = Tempearture

Since the volume and the moles of the gas are equal at both the temperatures, therefore we can rewrite the above eq. as:

$\boxed{\bold{\frac{P_1}{T_1} = \frac{P_2}{T_2} }}$  ⇒ Gay-Lussac's Law

Here we have,

P₁ = initial pressure of helium = 1.03 atm

P₂ = final pressure of helium = 2.29 atm

T₂ = Final temperature = 37.1 °C = 37.1 + 273 K = 310.1 K

T₁ = initial temperature

Now, we will substitute the values in Gay-Lussac's law to find the value of T₁.

∴ $\frac{1.03}{T_1} = \frac{2.29}{310.1}$

⇒ $T_1 = \frac{1.03\:\times\:310.1}{2.29}$

⇒ $T_1 = \frac{319.403}{2.29}$

⇒ $\bold{T_1 = 139.47\:K}$

Thus, the initial temperature of the gas in the cylinder was 139.47 K.

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