eozoischgc

2022-01-17

An aerosol spray can with a volume of 250 mL contains 2.30 g of propane gas (C3H8) as a propellant.
If the can is at ${25}^{\circ }C$, what is the pressure in the can?
What volume would the propane occupy at STP?
The can says that exposure to temperatures above 130 ∘F may cause the can to burst. What is the pressure in the can at this temperature?

ambarakaq8

Expert

Step 1
Given, Volume $=250mL$
Mass of propane gas $=2.30g$.
The number of moles of propane gas can be calculated as follows:
moles of propane $=\frac{mass}{\text{molar mass}}$
$=\frac{2.30g}{44.1\frac{g}{m}ol}$
$=0.0522mol$
Step 2
The pressure in the can is calculated by ideal gas equation as follows:
Temperature $={25}^{\circ }C=298K$
Ideal gas equation:
$PV=nRT$
$P=\frac{nRT}{V}$

$=5.11atm$
Step 3
The volume of propane occupy at STP can be calculated as follows:
At STP:

${T}_{1}=273K$
${V}_{1}=?$
GIven conditions:
${P}_{2}=5.11atm$
${T}_{2}=298K$
${V}_{2}=0.250L$
$\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}$

$\frac{{V}_{1}}{273}=0.00429$
${V}_{1}=0.00429×273$
$=1.171L$
$=1171mL$
Step 4
The pressure in the can at temperature of ${130}^{0}F$ can be calculated as
Temperature $={130}^{0}F={54.4}^{0}C=327.4K$
Ideal gas equation:
$PV=nRT$
$P=\frac{nRT}{V}$

$=5.59atm$

Toni Scott

Expert

moles propane
From
At STP 1 mole of a gas
AT ${130}^{\circ }F=403K$
$P=nR\frac{T}{V}=\left(0.052\right)\left(0.0821\right)\frac{403}{0.25}=6.9atm$

alenahelenash

Expert

Step 1 a) moles of propane $=\text{mass / molecular mass}=2.3/44=0.0523=n$ Volume $=V=250ml=0.25L$ $T={23}^{\circ }C=\text{Temperature}=273+23=296K$ Since $PV=nRT$ where $R=0.0821$ Hence substituting the values will give $P×0.25=0.0523×0.0821×296$ Step 2 b) Since at STP, $T=273K$ moles calculated above $=n=0.0523$ Hence substituting the values in $PV=nRT$ $⇒1×V=0.0523×0.0821×273$ $⇒V=1.17L$ Step 3 c) Since moles $=n=0.0523$ Volume of can $=250ml=0.25L$ $T={130}^{\circ }F$ Since the relation between Temperature in ${F}^{\circ }$ and K is given by $\left({F}^{\circ }-32\right)×\frac{5}{9}+273=K$ Hence, $\left(130-32\right)×5/9+273=327.444K=T$ Hence substituting the values in $PV=nRT$ we get $P×0.25=0.0523×0.0821×327.444$ .