# Question #7b4e4

Apr 27, 2014

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

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

${V}_{1}$ = 10.1 L

${T}_{1}$ = 273 + 25 = 298 K

n = mass / molar mass = 1.25 g / 20.1 g$m o {l}^{-} 1$ = 0.062 mol

${P}_{1}$ 10.1 L = 0.062 mol x 298 K x 8.314 J ${K}^{-} 1$ $m o {l}^{-} 1$

${P}_{1}$ = 0.062 mol x 298 K x 8.314 J ${K}^{-} 1$ $m o {l}^{-} 1$ / 10.1 L

${P}_{1}$ = 15.20 atm

We can find the effect of temperature on the pressure using the P-T law.

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

15.20 atm / 298 K = ${P}_{2}$ / 323K

15.20 atm x 323 K / 298 K = ${P}_{2}$

${P}_{2}$ = 4909.6 atm K / 298 K = 16.47 atm