## Calculate the volume of gas under standard conditions

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### Calculate the volume of gas under standard conditions

Biogas is also known as biogas. Under standard conditions, Biogas contains 60% methane by volume and the remaining is carbon dioxide and other gases (knowing that 1 mole of gas under standard conditions occupies a volume of 24.79 liters). A gas tank (liquefied petroleum gas) contains a mixture of propane and butane with a molar ratio of 1: 2. When completely burned, 1 mole of propane gives off 2220 kJ of heat, 1 mole of butane gives off 2850 kJ of heat. and 1 mole of methane gives off 890.5 kJ of heat. On average, a household needs to use up a 12kg gas cylinder every 60 days (assuming all reactions occur at 100%). After constructing and installing a replacement Biogas tank, what is the minimum volume of Biogas (under standard conditions) that must be generated in 60 days?

- ChenBeier
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### Re: Calculate the volume of gas under standard conditions

Molar mass propane C

12 kg is then 75 times more means 3300g propane and 8700 g butane.

In mol its 75 : 150 mol

Energy is 75 mol x 2220 kJ/ mol = 166500 kJ and 150 mol x 2850 kJ/mol = 427500 kJ in sum its 594000 kJ

In comparison to methane with 890,5 kJ/ mol we have to use 667,04 mol then.

1 mol = 24,79 l means we need 16535,95 l = 16,54 m^3

The methane is 60 % means we need 27,56 m^3 gas tank.

_{3}H_{8}= 44 g/mol and butane C_{4}H_{10}= 58 g/mol ratio 1:2 means 44 g : 116 g, in sum 160 g12 kg is then 75 times more means 3300g propane and 8700 g butane.

In mol its 75 : 150 mol

Energy is 75 mol x 2220 kJ/ mol = 166500 kJ and 150 mol x 2850 kJ/mol = 427500 kJ in sum its 594000 kJ

In comparison to methane with 890,5 kJ/ mol we have to use 667,04 mol then.

1 mol = 24,79 l means we need 16535,95 l = 16,54 m^3

The methane is 60 % means we need 27,56 m^3 gas tank.