Quantity of the substance, moles
4.4
Molar mass of the substance
18.0
Start with the ideal gas equation, <span><span><span>PV=nRT</span> </span><span>PV=nRT</span></span>
and rearrange for pressure to get <span><span><span>p=<span><span>nRT</span>V </span></span> </span><span>p=<span><span>nRT</span>V</span></span></span>
. You have all the necessary variables in their proper units, so plug em' into the equation to solve for pressure in units of atmospheres.
<span><span><span>P=<span><span>(2.5 mol)(300 K)(0.08206 L atm <span><span>K </span><span><span>−1</span> </span></span> mo<span><span>l </span><span><span>−1</span> </span></span></span><span>50.0 L</span> </span>=1.23 atm</span> </span><span>P=<span><span>(2.5 mol)(300 K)(0.08206 L atm <span>K<span>−1</span></span> mo<span>l<span>−1</span></span></span><span>50.0 L</span></span>=1.23 atm</span></span>
All that needs to be done now is converting atmospheres to mm <span><span><span>Hg</span> </span><span>Hg</span></span>
.
<span><span><span>1.23 atm∗<span><span>760 mm Hg</span><span>1 atm</span> </span>=935 mm Hg</span> </span><span>1.23 atm∗<span><span>760 mm Hg</span><span>1 atm</span></span>=935 mm Hg</span></span>
.
That value makes sense, since the original pressure in atmospheres was above 1, the pressure in mm <span><span><span>Hg</span> </span><span>Hg</span></span>
will be above 760.
Answer:
0.39 mol
Explanation:
Considering the ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
At same volume, for two situations, the above equation can be written as:-
Given ,
n₁ = 1.50 mol
n₂ = ?
P₁ = 3.75 atm
P₂ = 0.998 atm
T₁ = 21.7 ºC
T₂ = 28.1 ºC
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (21.7 + 273.15) K = 294.85 K
T₂ = (28.1 + 273.15) K = 301.25 K
Using above equation as:
Solving for n₂ , we get:
n₂ = 0.39 mol