# What is the volume of 2.0 moles of an ideal gas at 290 K and 2.8 atm?

Jan 31, 2016

$0.016996651 {m}^{3}$

#### Explanation:

here,

$t e m p e r a t u r e , T = \textcolor{red}{290 K}$

$p r e s s u r e = 2.8 a t m = 2.8 \times 101325 p a = \textcolor{red}{283710 p a}$

$n = \textcolor{red}{2.0 m o l e s}$

for ideal gas, we know,

$P V = n R T$

$V = \frac{n R T}{P}$

V=(2.0mol exx8.314Nmmol e^(-1)K^(-1)xx290K)/(283710Nm^(-2)

$= 0.016996651 {m}^{3}$