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what is the volume in milliliters of 0.0100 mol of CH(subscript4) gas at STP?

could not figure out how to add the 4 subscript for CH

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Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
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What is the volume (in mL) of 0.0100 moles of CH4 gas at STP?

Here, we know to use the ideal gas law equation, which relates the volume, temperature, and pressure of an ideal gas:     PV=nRT

where     P = the pressure of the gas in atmospheres (atm)

              V = the volume of the space the gas is contained in in liters (L)

              n = the number of moles of the gas (mol) = 0.0100 mol

              R = the universal gas constant, 0.0821 L·atm/K·mol  

              T = the absolute temperature of the gas in Kelvins (K)

A gas at standard temperature and pressure (STP) is defined at a temperature of 0°C and at a pressure of 1 atm.

Since the the ideal gas law requires the absolute temperature of a gas, we have to convert the temperature of the gas at STP from °C to Kelvins:  

        K = °C + 273.15     ==>     K = 0°C + 273.15 = 273.15 K

Sine we are solving for the volume of the gas, we take the equation PV=nRT and divide both sides by P to solve for V:

                  V = nRT/P

 V = [(0.0100 mol)*(0.0821 L·atm/K·mol)*(273.15 K)] / (1 atm) = 0.22425615 L ≈ 0.22426 L

         (0.22426 L) * (1000 mL / 1 L) = 224.26 mL

Daniel O. | Math and Physics Tutor, with a math and physics degreeMath and Physics Tutor, with a math and ...
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To use subscript, use the x2 button: CH4

We know PV = nRT, so we want V = nRT/P  where R = 8.314 J/(Mol K)

At STP, P = 100 kPa = 100,000 Pa and T = 273.15 K 

n = 0.01 mol

So substitute it all into the equation for V = 0.01 * 8.314 * 273.15 / 100,000

V = 0.0002271 = 2.271 x 10-4 m3 = 0.2271 L


You can also use the fact that 1 mol of gas at STP = 22.41L, and then multiply that by the number of moles n = 0.01

0.01 * 22.41L = 0.2241 L

CHARLES R. | Get Expert Help in Chemistry, Algebra, Trigonometry, Computers, ExcelGet Expert Help in Chemistry, Algebra, T...
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The question is more easily solved if you remember that at STP, 1 mole of gas occupies 22.4 L of volume. Thus 0.0100 mole would occupy 1/100 th of the 22.4 L or 0.224 L. To convert the answer to mL, just multiply the 0.224 L by the unit factor of 1000 mL / 1 L to get a value of 224 mL.


I love using the STP trick! It's important to understand how and when you use PV=nRT, but if they give you an unknown mass at STP, why go through all of the extra steps (which gives the opportunity for mistakes)? Chances are using the 1 mol @ STP method is what they want you to do for this question, and all they'll give you time to do for a question like this on the exam.

Tara B. | Math and Science Tutor Available NowMath and Science Tutor Available Now

OK, so Daniel and Tamara got slightly different answers.  Why is that?  And which one is correct?

They both use the ideal gas law: PV=nRT.  And they isolated V: V=nRT/P  This is correct.

They both agree on the number of moles of gas, which is given in the problem:

n = the number of moles of the gas (mol) = 0.0100 mol

They give different values for they ideal gas constant, R:

Tamara says R = the universal gas constant = 0.0821 L·atm/K·mol

Daniel uses R = 8.314 J/(mol K)

They are the same value, just with different units.  If you don't believe me, just convert them to the same units.

Where they disagree is in their defnintion of STP.

They both use a temperature of T=273.15K, but Tamara says the pressure is P=1 atm, and Daniel says it is P=100kPa.  These are not the same value.  1 atm is equal to 101.325 kPa, not 100kPa.

The reason for this is that the definition of STP has been changed by IUPAC.  STP, is defined by IUPAC as a pressure of 100 kPa which is equal to 0.9869 atm and a temperature of 273.15 K.  It used to be 1atm, but IUPAC changed it to .9869 atm in 1982 (10/28/12

So Daniel was correct.  We must use 100kPa or .9869 atm as the pressure:

P =pressure of the gas = .9869 atm

Now all there is left to do is substutite your known values for P, n, R, and T into the ideal gas equation, and you can see if Daniel got the correct answer using the proper value of P.

Is Daniel's answer perfect?

Rememer, your solution can have only as many significant figures as your knowns have.  Can Daniel know for sure that the volume is .2271 given that the known n=.0100 only has three significant figures?  Be careful not to report too many significant figures because it implies that you know the answer more precisely than you do.

Also, remember to report the answer in the units the question is asking for.

Helpful tip for all chemistry and physics problems:

It is very good how Tamara included the units in her calculations.  The only way you can know for sure if you were using the right units is if they all cancel out by the end, and you end up with the units you are supposed to have.  When you write it out, it helps to draw a line and write all the numerators on top and the denominators on bottom.  For example, I would write R like this:

0.0821 L·atm

---------------------   *   


That way, you can easily see what units cancel out.


I agree that it's a good idea to include units in your calculations in chem/physics to make sure you cancel out all the correct units (it's just much easier to do when writing by hand, than typing it all in, hence why i left it out). Though, if you use all the most standard units, your result will always have the correct units (provided you know what that resulting unit is), but if you're unsure, it's better to write them all in.

You're right about significant figures - it's been a few years since I've done chemistry, and I just couldn't remember if .0100 counted as four or three S.F.'s.

You're right.  It's definitely more cumbersome to try to type equations than to write them with pen and paper.

The reason .0100 counts as three s.f.'s rather than four is because leading zeros don't count.  The leading zero doesn't give you any extra information.