# Question #85a0f

Apr 27, 2014

To solve this problem , we will need to use the ideal gas equation;

${P}_{1} {V}_{1}$= nR${T}_{1}$

${V}_{1}$ = 456 mL 0r 0.456 L

Standard Temperature is 0 degree Celsius or 273 K

${T}_{1}$ = 273 K

Standard pressure is 1 atm

${P}_{1}$ = 1 atm

${P}_{2}$ = 1.22 atm , ${V}_{2}$ = ? ${T}_{2}$ = 273 + 112 = 385 K

${P}_{1}$ x ${V}_{1}$ / ${T}_{1}$ = ${P}_{2}$ x ${V}_{2}$ / ${T}_{2}$

1 atm x 0.456 L / 273 K = 1.22 atm x ${V}_{2}$ / 385 K

Cross multiplying

${V}_{2}$ = 1 atm x 0.456 L x 385 K / 273 K x 1.22 atm

${V}_{2}$ = 0.527 L 0r 527 mL.