Question

A gas with volume of 700.0 mL at STP (0.0 0C, 1.00 atm) is compressed to a volume of 200.0 mL and the temperature is increased to 30.0 0C. What is the new pressure in kPa?

Answer 1

393.1 KPa

Step-by-step explanation:

Now we have to use the general gas equation;

P1V1/T1 = P2V2/T2

P1= initial pressure of the gas= 1.00atm

V1 = initial volume= 700.0 ml

T1= initial temperature= 0.0°C +273= 273K

V2= final volume= 200.0 ml

T2= final temperature =30.0°C +273= 303 K

P2= final pressure =????

Therefore;

P2= P1V1T2/ V2T1

P2= 1.00 × 700.0 × 303/ 200.0 × 273

P2= 3.88 atm

But

1 atm = 101.325 kilopascals

Therefore 3.88 atm = 3.88 × 101.325 kilopascals = 393.1 KPa