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

The correct answer is a