[acolegario2]

Hi. I wonder if you could help me with this exercise:

1) The owner of a house uses 4 x 10³ m³ of natural gas in a year to heat his home. Suppose that the gas natural is mainly methane, CH₄, and the methane is a gas perfect for the conditions of this problem, which are 1.00 atm and 20 ° C. What is the mass of gas used?

[acolegario2]

Now show your attempt.

You mentioned the Ideal Gas law, do you have the formula?  Can you write it for us?  Can you substitute the values given in the problem?

Sorry, i was working with the exercise, i tried with this formula:



I obtained a different result from the answer that i got. Answer is: 2,67 Mg.

[Borek]

Sorry, i was working with the exercise, i tried with this formula:

You are on the right track, but you have ignored units. You got some number and you assumed it is the mass you are looking for. It is not. You calculated n, what it is?

Plus, check your math, as the number you got is off by the factor of 10.

[AWK]

pV=nRT => (n=m/M) => pV=mRT/M => m=pVM/(RT) These are elementary algebraic operations (substitution and formula transformation).
Just put correct units into formula (units should be canceled, except mass in g , kg or Mg)
Answer in your textbook is wrong for two reasons - wrong number and wrong number of significant figures (if your data are correct).

[acolegario2]

You are on the right track, but you have ignored units. You got some number and you assumed it is the mass you are looking for. It is not. You calculated n, what it is?

Plus, check your math, as the number you got is off by the factor of 10.

pV=nRT => (n=m/M) => pV=mRT/M => m=pVM/(RT) These are elementary algebraic operations (substitution and formula transformation).
Just put correct units into formula (units should be canceled, except mass in g , kg or Mg)
Answer in your textbook is wrong for two reasons - wrong number and wrong number of significant figures (if your data are correct).

Thanks for your help. I applied your corrections in the formula and got this result:



Yes, maybe the answer in the text i guess is wrong with Mg. unit.

[Borek]

Answer in your textbook is wrong for two reasons - wrong number and wrong number of significant figures

2.67 Mg is a perfectly correct number (you are right about significant figures though, but reporting the answer with a single digit would look a bit funny to me).

[Borek]

No, the result is not 2,66 kg. Again you have ignored the units.

What was the EXACT number you got?

[acolegario2]

No, the result is not 2,66 kg. Again you have ignored the units.

What was the EXACT number you got?

I got 2.668561461

[AWK]

May fault. I took M of water instead and got result very close to 3.

You used kmol in R constant and molecular mass in kg, but for mol.

[Enthalpy]

Methane is about half as light as air which weighs a bit more than 1kg/m3
so
4000m³ methane must weigh around 2000kg, not 2kg.

Be also careful with the decimal separator: "." in English (or at least in the UK and US).

Pure chemists seem happy with grams, but if you plan to make also mechanical and acoustical engineering, I recommend to opt for Si units once for good, with kg and Pa. Whatever your choice, always check for g versus kg (and versus N).

[acolegario2]

Methane is about half as light as air which weighs a bit more than 1kg/m3
so
4000m³ methane must weigh around 2000kg, not 2kg.

Be also careful with the decimal separator: "." in English (or at least in the UK and US).

Pure chemists seem happy with grams, but if you plan to make also mechanical and acoustical engineering, I recommend to opt for Si units once for good, with kg and Pa. Whatever your choice, always check for g versus kg (and versus N).

May fault. I took M of water instead and got result very close to 3.


You used kmol in R constant and molecular mass in kg, but for mol.

No, the result is not 2,66 kg. Again you have ignored the units.

What was the EXACT number you got?

Yes, in class we use SI units system. At the end i don't know which unit i ignored? How do i get Mg unit?

[AWK]

R [m³·kpa/(K·kmol)]
M [kg/mol]

[Borek]

Note: Mg is not an SI unit, but is close to the system and  trivial to convert. M stands for Mega (as in SI), g stands for gram. Not very different form kg conceptually.

[AWK]

Excerpt from NIST SI Guide"
=symbols for decimal multiples and submultiples of the unit of mass are formed by attaching SI prefix symbols to g, the unit symbol for gram, and the names of such multiples and submultiples are formed by attaching SI prefix names to the name "gram."=
i.e.
1 mg = 10^-6 kg
1 Mg = 10⁶ g =10³ kg corrected

[Borek]

1 Mg = 10⁶ kg

Nope. Mg is not 10⁶ kg.

This is tricky as technically g is not an SI unit. kg - which name suggests it is already a multiple of a basic unit - is.