# What volume would a 10*g mass of carbon monoxide enclosed in a piston occupy at 1*atm, and 0 ""^@C?

Nov 28, 2016

Well, one mole of ideal gas occupies 22.4 L.

#### Explanation:

So, we have $\frac{10 \cdot g}{28.01 \cdot g \cdot m o l} \cong \frac{1}{3} \cdot \text{mole}$.

And thus if we assume carbon monoxide behaves ideally, its volume (if enclosed in a piston expanding against atmospheric pressure) would be approx. $7.4 \cdot L$.

Nov 29, 2016

8.00 Liter

#### Explanation:

It can be further explained as :-

idle gas equation is

$P V = n R T$

Here ,

P = 1 atm
V = To be find out
R = 0.08206 Universal Gas constant
T = 273 k
$n = \frac{m}{M}$
m = mass of gas [g], given 10 g
M = mass of 1 mole of gas [g] that is 28.01 of CO

By putting all values -

$1 \cdot V = \frac{10}{28.01} \cdot 0.08206 \cdot 273$
or
$V = \frac{24.238}{28.01}$
or
V = 8.00

Hence Answer is 8.00 Liter Volume