Answer: The volume of the container needed is 554.6 L
Explanation:
To calculate the volume of the gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 3.0 atm
V = Volume of the gas = ? L
T = Temperature of the gas = [tex]25^oC=[25+273]K=298K[/tex]
R = Gas constant = [tex]0.0821\text{ L.atm }mol^{-1}K^{-1}[/tex]
n = number of moles of propane gas = 68.0 moles
Putting values in above equation, we get:
[tex]3.0atm\times ?L=68mol\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\V=\frac{68\times 0.0821\times 298}{3.0}=554.6L[/tex]
Hence, the volume of the container needed is 554.6 L
The volume of the container needed is 554.6 L
To compute the volume of the gas, we use the ideal gas equation, which is as follows:
[tex]PV=nRT[/tex] ....(1)
Here,
P = pressure of the gas = 3.0 atm
V = Volume of the gas = ? L
T = Temperature of the gas =25° C=(25+273)K=298 K
R = Gas constant = 0.0821 Latm/molK
n = number of moles of propane gas = 68.0 moles
Substitute all the values in the equation (1) as follows:-
[tex]3\ atm\times V=68\ mol\times0.0821\ Latm/Kmol\times 298\ K\\\\V=\frac{68\ mol\times0.0821\ Latm/Kmol\times 298\ K}{3.0\ atm}\\\\= 554.6\ L[/tex]
Hence, the volume of the container needed to hold the same amount of propane is 554.6 L
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