Question

A mylar balloon is filled with helium gas to a pressure of 107 kPa when the temperature is 22°C. If the temperature changes to 45°C, what will be the pressure of the helium in the balloon?

Answer 1

Answer:

Explanation:

To solve this problem, we have to remember the ideal gas theory. We going to use the Gay Lussac Law.

The equation is:

PiTo = PoTi

Where:

Pi =initial pressure

Po = final pressure

Ti = initial temperature

To = final temperature

Let's use this equation and replace the values:

Pi = 107 kPa

Ti = 22°C

To = 45 °C

Po = ?

You can use the units as the problem gave you.

Using the Gay Lussac law:

107 kPa x 45 °C = Po x 22°C

Po = (4,815 kPa°C / 22°C) = 218.86 kPa

Po = 218 kPa

Answer 2

The pressure of helium in the balloon will be 115.34 kPa.

To calculate the final pressure of helium in the balloon, we use the formula below.

Formula:

• P/T = P'/T'............... Equation 1

Where:

• P = Initial pressure of helium in the balloon
• T = Initial temperature of helium in the balloon
• P' = Final pressure of helium in the balloon
• T' = Final temperature of helium in the balloon.

Make P' the subject of the equation

• P' = PT'/T................ Equation 2

From the question,

Given:

• P = 107 kPa
• T = 22°C = (22+273) = 295 K
• T' = 45°C = (45+273) = 318 K

Substitute the given values into equation 2

• P' = (107×318)/295
• P' = 115.34 kPa.

Hence, The pressure of helium in the balloon will be 115.34 kPa.
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