An lpg cylinder contains 15kg of butane at 27oC and 10 bar pressure. The cylinder leaked and it was found that after 4 days,the mass of the gas left in cylinder in was only 10 kg. How much pressure was dropped in each day (average pressure drop/day)?

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Solution

Dear Student,

According to ideal gas equation, PV = nRT
In the question, V, R and T remains constant. Only P and n are changing
P is directly proportional to n.
n = number of moles = mass/molar mass
Molar mass in both the cases is also constant, so. P is directly proportional to mass of gas.
$\frac{P_{1}}{P_{2}} = \frac{m_{1}}{m_{2}}$
P₁ = 10 bar, P₂ = ?
m₁ = 15 kg, m₂ = 10 kg
After putting the values,
$\frac{10}{P_{2}} = \frac{15}{10}$
P₂ = 6.67 bar
Pressure loss = 10-6.67 = 3.33 bar
Number of days = 4
Pressure drop per day = 3.33/4 = 0.83 bar/day

Regards

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