A gas cylinder contains 14.2 kg butane gas; it is consumption of energy in a family is 10000 KJ for cooking purposes how long this cylinder will be sufficient to supply. If enthalpy of combustion of butane is 2685 KJ mol.
Answer
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Hint: Butane is having the formula ${{C}_{4}}{{H}_{10}}$, it is actually found as gas at room temperature. It is a very dangerous, flammable substance. Butane gas is colourless, and is found to have many wide applications. We will first find the molecular weight of butane, then heat released during combustion of butane, by the formula:
\[\dfrac{enthalpy\text{ }of\text{ }fusion}{molecular\text{ }weight}\times amount\text{ }of\text{ }gas\]
Complete step by step answer:
We have to find the time here that is how long this cylinder will be sufficient to supply, for calculating it we are being provided with following data:
- Enthalpy of combustion of butane is given=2658KJ/mol,
- We are given the amount of butane gas in the cylinder= 14.2 kg, we will convert it into grams, that is =14200g
- Molecular weight of butane=58gm/mol, we calculated it as, ${{C}_{4}}{{H}_{10}}$=
\[\begin{align}
& 4\times 12+10\times 1 \\
& =48+10 \\
& =58 \\
\end{align}\]
- enthalpy of combustion of butane or we can say the heat released during combustion of 58g of butane =2658 KJ
-So, we will find the heat released during the combustion of 14200g of butane=
\[\dfrac{enthalpy\text{ }of\text{ }fusion}{molecular\text{ }weight}\times amount\text{ }of\text{ }gas\]
$\dfrac{2658}{58}\times 14200\text{ }KJ$
- We know that 10000 KJ of energy is consumed in one day, therefore, the number of days require to consume $\dfrac{2658}{58}\times 14200\text{ }KJ$of energy will be equal to :
\[\begin{align}
& \dfrac{1}{10000}\times \dfrac{2658}{58}\times 14200 \\
& =0.0001\times 45.8275\times 14200 \\
& =65.075 \\
\end{align}\]
= 65 days
- Hence, we can say that the cylinder will be sufficient for 65 days.
Additional Information
- It is found that when butane gas is mixed with other hydrocarbons, it may be called as LPG, it is also used as a petrol component. It has great demand as a cigarette lighter and also used in deodorants as well. Hence, we can say that it is economically very much useful.
Note:
- We must remember to convert the amount of gas given in kilograms into grams, and then solve the question.
- As butane gas is highly flammable so one should take all the precautions while handling or we can say using this gas.
\[\dfrac{enthalpy\text{ }of\text{ }fusion}{molecular\text{ }weight}\times amount\text{ }of\text{ }gas\]
Complete step by step answer:
We have to find the time here that is how long this cylinder will be sufficient to supply, for calculating it we are being provided with following data:
- Enthalpy of combustion of butane is given=2658KJ/mol,
- We are given the amount of butane gas in the cylinder= 14.2 kg, we will convert it into grams, that is =14200g
- Molecular weight of butane=58gm/mol, we calculated it as, ${{C}_{4}}{{H}_{10}}$=
\[\begin{align}
& 4\times 12+10\times 1 \\
& =48+10 \\
& =58 \\
\end{align}\]
- enthalpy of combustion of butane or we can say the heat released during combustion of 58g of butane =2658 KJ
-So, we will find the heat released during the combustion of 14200g of butane=
\[\dfrac{enthalpy\text{ }of\text{ }fusion}{molecular\text{ }weight}\times amount\text{ }of\text{ }gas\]
$\dfrac{2658}{58}\times 14200\text{ }KJ$
- We know that 10000 KJ of energy is consumed in one day, therefore, the number of days require to consume $\dfrac{2658}{58}\times 14200\text{ }KJ$of energy will be equal to :
\[\begin{align}
& \dfrac{1}{10000}\times \dfrac{2658}{58}\times 14200 \\
& =0.0001\times 45.8275\times 14200 \\
& =65.075 \\
\end{align}\]
= 65 days
- Hence, we can say that the cylinder will be sufficient for 65 days.
Additional Information
- It is found that when butane gas is mixed with other hydrocarbons, it may be called as LPG, it is also used as a petrol component. It has great demand as a cigarette lighter and also used in deodorants as well. Hence, we can say that it is economically very much useful.
Note:
- We must remember to convert the amount of gas given in kilograms into grams, and then solve the question.
- As butane gas is highly flammable so one should take all the precautions while handling or we can say using this gas.
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