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?

393.1 KPa

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