Question

Given a gas whose temperature is 418 K at a pressure of 56.0 kPa. What is the pressure of the gas if its Temperature changes to 64°C?

Answer 1

Answer: P₂=0.44 atm

Explanation:

For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.

Gay-Lussac's Law: $\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }$

First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.

64+273K=337K

Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.

$56.0kPa*\frac{1000Pa}{1kPa} *\frac{atm}{101325Pa} =0.55atm$

Since our units are in atm and K, we can use Gay-Lussac's Law to find P₂.

$P_{2} =\frac{T_{2} P_{1} }{T_{1} }$

$P_{2}=\frac{(337K)(0.55atm)}{418K}$

P₂=0.44 atm