Answer and Explanation:
We have given the expression
Dimension of
![\rho =ML^(-3)](https://img.qammunity.org/2020/formulas/physics/college/jhlhjl4shz6gmcemwcyif1svej5d0gm2ln.png)
Dimension of R which is gas constant
![=ML^2T^(-2)\Theta ^(-1)M^(-1)](https://img.qammunity.org/2020/formulas/physics/college/9altpci9kwcsf0jxfobs6cgu215eww6cnn.png)
Dimension of temperature T
![\Theta ^(-1)](https://img.qammunity.org/2020/formulas/physics/college/5pgz2ozz6sk9kpxlw0pvapopevxqu0xt33.png)
And dimension of pressure
![ML^(-1)T^(-2)](https://img.qammunity.org/2020/formulas/physics/college/9hs0s0mv8j6rjzvhnnt750bb3subllhp4y.png)
Now combine dimension of
![=ML^(-3)ML^2T^(-2)\Theta ^(-1)M^(-1)\Theta ^(-1)=ML^(-1)T^(-2)](https://img.qammunity.org/2020/formulas/physics/college/hamp7xp51k1htqfxzagt53dtqk47yju0ft.png)
So the dimension of
and dimension P is same so there unit will also be same
From ideal gas equation we know that
![PV=nRT](https://img.qammunity.org/2020/formulas/chemistry/high-school/uelah1l4d86yyc7nr57q25hwn1eullbhy3.png)
![P=(n)/(V)RT=\rho RT](https://img.qammunity.org/2020/formulas/physics/college/ixt9tyn28tyuh77iie33xs8bxmwfc1v0ia.png)
As the both P and
has same dimension so they are dimensionally constant