If 20 ml of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm what's the best expression representing the new volume

V1 = 20 ml=0.020 L

T1 = 10⁰+273⁰ = 283 K

T2 = 100⁰ + 273⁰ = 373 K

P1 = 1 atm

P2 = 10 atm

V2-?

P1V1/T1 = P2V2/T2

(1 atm * 0.020 L) / 283K = 10 atm*V2/373K

V2 = (1 atm * 0.020 L*373K) / (283K * 10 atm) =

(1 * 0.020 * 373) / (283 * 10) = 0.00264 L ≈ 2.6 ml