# Does Gas constant depend on molecular weight?

I came across the following question recently

Calculate the difference between two specific heat of 1 g of helium gas at NTP. Molecular weight of helium = 4 and J = $$4.186×10^7$$ erg $$cal^{-1}$$

The solution given is

If R is called the “Universal” Gas constant why does it change for different molecular masses?

I found this website which talked about Gas constant vs. universal gas constant, but all the problems I’ve come across I’ve always used PV=nRT and haven’t had to consider the molecular weight. Why is this situation any different?

Could you please point out where I’m making a mistake or having a misconception on this topic?

but all the problems I’ve come across I’ve always used PV=nRT and haven’t had to consider the molecular weight. Why is this situation any different?

Because $$R$$ in the equation $$PV=nRT$$ is the universal gas constant, sometimes designated with an overhead bar as $$\bar R$$, is a single value that can be applied any ideal gas when the amount of the gas is given in moles ($$n$$).

On the other hand, for the equation

$$PV=mR_{g}T$$

where $$m$$ is the mass of the gas, $$R_{g}$$ is the specific gas constant. Since the number of moles of gas equals its mass divided by its molecular weight, the relationship between $$\bar R$$ and $$R_g$$ is

$$R_{g}=\frac{\bar R}{mol.wt}$$

Hope this helps.

• Just for clarification, “where m is the mass of the gas”, in the equation m is the mass? Not molecular weight? May 9, 2022 at 15:37
• @Natru Correct. $m$ is the mass, not the Molecular Weight. May 9, 2022 at 16:49
• So, nR = mR_g like my downvoted answer ;) May 10, 2022 at 9:52
• @buddhabrot yes. Have no idea why your answer was downvoted May 10, 2022 at 10:07

$$nR_{universal} = mR_{specific}$$, so if you deal with amount of molecules you can use the universal gas constant in the ideal gas law.

• Not sure why anyone downvoted this May 9, 2022 at 9:44