An aerosol spray can with a volume of 250 mL contains 2.30 g of propane...

Step 1
Given, Volume ${\displaystyle =250mL}$
Mass of propane gas ${\displaystyle =2.30g}$.
The number of moles of propane gas can be calculated as follows:
moles of propane ${\displaystyle =\frac{mass}{\text{molar mass}}}$
${\displaystyle =\frac{2.30g}{44.1\frac{g}{m}ol}}$
${\displaystyle =0.0522mol}$
Step 2
The pressure in the can is calculated by ideal gas equation as follows:
Temperature ${\displaystyle ={25}^{\circ }C=298K}$
Ideal gas equation:
${\displaystyle PV=nRT}$
${\displaystyle P=\frac{nRT}{V}}$
${\displaystyle =\frac{0.0522mol\times 0.0821Latm{K}^{-1}mo{l}^{-1}\times 298K}{0.250L}}$
${\displaystyle =5.11atm}$
Step 3
The volume of propane occupy at STP can be calculated as follows:
At STP:
${\displaystyle P_1=1atm}$
${\displaystyle T_1=273K}$
${\displaystyle V_1=?}$
GIven conditions:
${\displaystyle P_2=5.11atm}$
${\displaystyle T_2=298K}$
${\displaystyle V_2=0.250L}$
${\displaystyle \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}}$
${\displaystyle \frac{1atm\times V_1}{273K}=\frac{5.11atm\times 0.250L}{298K}}$
${\displaystyle \frac{V_1}{273}=0.00429}$
${\displaystyle V_1=0.00429\times 273}$
${\displaystyle =1.171L}$
${\displaystyle =1171mL}$
Step 4
The pressure in the can at temperature of ${\displaystyle {130}^{0}F}$ can be calculated as
Temperature ${\displaystyle ={130}^{0}F={54.4}^{0}C=327.4K}$
Ideal gas equation:
${\displaystyle PV=nRT}$
${\displaystyle P=\frac{nRT}{V}}$
${\displaystyle =\frac{0.0522mol\times 0.0821Latm{K}^{-1}mo{l}^{-1}\times 327.4K}{0.250L}}$
${\displaystyle =5.59atm}$

moles propane $=2.30g\times 1mole/44g=0.052moles$
From ${\displaystyle PV=nRT,P=nR\frac{T}{V}=\left(0.052\right)\left(0.0821\right)\frac{298}{0.25}=5.1atm}$
At STP 1 mole of a gas $=22.4L,thus0.052moles\times 22.4L/mole=1.16liters$
AT ${\displaystyle {130}^{\circ }F=403K}$
${\displaystyle P=nR\frac{T}{V}=\left(0.052\right)\left(0.0821\right)\frac{403}{0.25}=6.9atm}$

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 $P\times 0.25=0.0523\times 0.0821\times 296$ $\Rightarrow P=\text{pressure in the can or pressure of gas}=5.08atm$ Step 2 b) Since at STP, $T=273K$ $P=1atm$ moles calculated above $=n=0.0523$ Hence substituting the values in $PV=nRT$ $\Rightarrow 1\times V=0.0523\times 0.0821\times 273$ $\Rightarrow V=1.17L$ 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 $(F^{\circ }-32)\times \frac{5}{9}+273=K$ Hence, $(130-32)\times 5/9+273=327.444K=T$ Hence substituting the values in $PV=nRT$ we get $P\times 0.25=0.0523\times 0.0821\times 327.444$ $\Rightarrow P=5.624atm$.