Ideal Gas Law: Solving for V with Constant R: Units L or m3?

[anisotropic]

When using the ideal gas law equation and solving for V, with constant R = 8.314 J·mol⁻¹·K⁻¹, what are the resultant units? L or m³?

I always remember it being in L, but wouldn't it be m³ based purely on the units being put into the equation?

Other units I'm using:

P: Pa (not kPa)
T: K

 

[Borek]

m³

Trust your units

For L one uses 8.314x10³

 

[Gokul43201]

For L one uses 8.314x10³

Alternatively, there's also the other commonly used unit system, seen in R=0.0821 L-atm/(K-mol)

 

[nrqed]

When using the ideal gas law equation and solving for V, with constant R = 8.314 J·mol⁻¹·K⁻¹, what are the resultant units? L or m³?

I always remember it being in L, but wouldn't it be m³ based purely on the units being put into the equation?

Other units I'm using:

P: Pa (not kPa)
T: K

what you need to know to answer the question is that $1 Pa = 1 N/m^2$ (Newton per meter squared) and that $1 N = 1 J/m$ so that $1 Pa = 1 J/m^3$ which shows that V comes out in $m^3$