Question

A gas in a sealed container has an initial pressure of 125 kPa at 25.0∘C. If the pressure is increased to 150.0 kPa, what will the new temperature be?

Answer 1

Answer:

$T'=30^{\circ} C$

Explanation:

Initial pressure, P = 125 kPa

Initial temperature, T = 25.0°C

If new pressure is increased to P' = 150.0 kPa

We need to find new temperature. According to Gay-Lussac's law,

P ∝ T

or

$\dfrac{P}{T}=\dfrac{P'}{T'}\\\\T'=\dfrac{P'T}{P}\\\\T'=\dfrac{150\times 25}{125}\\\\=30^{\circ} C$

So, the new temperature will be $30^{\circ} C$.