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

eozoischgc

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 25C, 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?

Answer & Explanation

ambarakaq8

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 =massmolar mass
=2.30g44.1gmol
=0.0522mol
Step 2
The pressure in the can is calculated by ideal gas equation as follows:
Temperature =25C=298K
Ideal gas equation:
PV=nRT
P=nRTV
=0.0522mol×0.0821L atm K1mol1×298K0.250L
=5.11atm
Step 3
The volume of propane occupy at STP can be calculated as follows:
At STP:
P1=1 atm
T1=273K
V1=?
GIven conditions:
P2=5.11atm
T2=298K
V2=0.250L
P1V1T1=P2V2T2
1 atm×V1273K=5.11atm×0.250L298K
V1273=0.00429
V1=0.00429×273
=1.171L
=1171mL
Step 4
The pressure in the can at temperature of 1300F can be calculated as
Temperature =1300F=54.40C=327.4K
Ideal gas equation:
PV=nRT
P=nRTV
=0.0522mol×0.0821L atmK1mol1×327.4K0.250L
=5.59atm
Toni Scott

Toni Scott

Expert

2022-01-19Added 32 answers

moles propane =2.30g×1mole/44g=0.052 moles
From PV=nRT,P=nRTV=(0.052)(0.0821)2980.25=5.1 atm
At STP 1 mole of a gas =22.4L,thus 0.052 moles×22.4L/mole=1.16 liters
AT 130F=403K
P=nRTV=(0.052)(0.0821)4030.25=6.9atm

alenahelenash

alenahelenash

Expert

2022-01-23Added 366 answers

Step 1 a) moles of propane =mass / molecular mass=2.3/44=0.0523=n Volume =V=250ml=0.25L T=23C=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 P=pressure in the can or pressure of gas=5.08 atm Step 2 b) Since at STP, T=273K P=1 atm 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=130F Since the relation between Temperature in F and K is given by (F32)×59+273=K Hence, (13032)×5/9+273=327.444K=T Hence substituting the values in PV=nRT we get P×0.25=0.0523×0.0821×327.444 P=5.624 atm.

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