eozoischgc

Answered

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?

If the can is at

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?

Answer & Explanation

ambarakaq8

Expert

2022-01-18Added 31 answers

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:

$PV=nRT$

$P=\frac{nRT}{V}$

$=\frac{0.0522mol\times 0.0821L\text{}atm\text{}{K}^{-1}mo{l}^{-1}\times 298K}{0.250L}$

$=5.11atm$

Step 3

The volume of propane occupy at STP can be calculated as follows:

At STP:

${P}_{1}=1\text{}atm$

${T}_{1}=273K$

${V}_{1}=?$

GIven conditions:

${P}_{2}=5.11atm$

${T}_{2}=298K$

${V}_{2}=0.250L$

$\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}$

$\frac{1\text{}atm\times {V}_{1}}{273K}=\frac{5.11atm\times 0.250L}{298K}$

$\frac{{V}_{1}}{273}=0.00429$

${V}_{1}=0.00429\times 273$

$=1.171L$

$=1171mL$

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:

$PV=nRT$

$P=\frac{nRT}{V}$

$=\frac{0.0522mol\times 0.0821L\text{}atm{K}^{-1}mo{l}^{-1}\times 327.4K}{0.250L}$

$=5.59atm$

Given, Volume

Mass of propane gas

The number of moles of propane gas can be calculated as follows:

moles of propane

Step 2

The pressure in the can is calculated by ideal gas equation as follows:

Temperature

Ideal gas equation:

Step 3

The volume of propane occupy at STP can be calculated as follows:

At STP:

GIven conditions:

Step 4

The pressure in the can at temperature of

Temperature

Ideal gas equation:

Toni Scott

Expert

2022-01-19Added 32 answers

moles propane

From

At STP 1 mole of a gas

AT

alenahelenash

Expert

2022-01-23Added 366 answers

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 $P\times 0.25=0.0523\times 0.0821\times 296$ $\Rightarrow P=\text{pressure in the can or pressure of gas}=5.08\text{}atm$ Step 2
b) Since at STP, $T=273K$ $P=1\text{}atm$
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
$P\times 0.25=0.0523\times 0.0821\times 327.444$
$\Rightarrow P=5.624\text{}atm$ .

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