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

Step-by-step explanation:

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

Gay-Lussac's Law:
`(P_(1) )/(T_(1) ) =(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*(1000Pa)/(1kPa) *(atm)/(101325Pa) =0.55atm`

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

`P_(2) =(T_(2) P_(1) )/(T_(1) )`

`P_(2)=((337K)(0.55atm))/(418K)`

P₂=0.44 atm