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

2 Answers
anor277
Nov 28, 2016

Well, one mole of ideal gas occupies 22.4 L.

Explanation:

So, we have #(10*g)/(28.01*g*mol)~=1/3*"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*L#.

Suresh Rawal
Nov 29, 2016

8.00 Liter

Explanation:

It can be further explained as :-

idle gas equation is

#PV=nRT#

Here ,

P = 1 atm
V = To be find out
R = 0.08206 Universal Gas constant
T = 273 k
#n = 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*V=10/28.01*0.08206*273#
or
#V = 24.238/28.01#
or
V = 8.00

Hence Answer is 8.00 Liter Volume

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