Question

PLEASE ANSWER ASAP AND SHOW WORK: A camping stove uses a 5.0 L propane tank that holds 68.0 moles of liquid C3H8. How large a container would be needed to hold the same amount pf propane as a gas at 25.0 ⁰C and a pressure of 3.0 atm?​

Answer 1

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:

$PV=nRT$

where,

P = pressure of the gas = 3.0 atm

V = Volume of the gas = ? L

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

n = number of moles of propane gas = 68.0 moles

Putting values in above equation, we get:

$3.0atm\times ?L=68mol\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\V=\frac{68\times 0.0821\times 298}{3.0}=554.6L$

Hence, the volume of the container needed is 554.6 L