The dimensions of universal gas constant are

[L^{2} M^{1} T^{- 2} K^{- 1}]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[L^{1} M^{2} T^{- 2} K^{- 1}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[L^{1} M^{1} T^{- 2} K^{- 1}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[L^{2} M^{2} T^{- 2} K^{- 1}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $[L^{2} M^{1} T^{- 2} K^{- 1}]$
Using ideal gas equation, $R = \frac{P V}{n T} = \frac{W}{n T} = \frac{M L^{2} T^{- 2}}{m o l K}$ $= [L^{2} M^{1} T^{- 2} K^{- 1}]$

Suggest Corrections

Similar questions

[M L^{2} T^{- 2} K^{- 1}]

f (x) = c o s (\frac{π}{x}), x \neq 0

f (x) = \frac{a_{2 k} x^{2 k} + a_{2 k - 1} x^{2 k - 1} + . . . . + a_{1} x + a_{0}}{b_{2 k} x^{2 k} + b_{2 k - 1} x^{2 k - 1} + . . . . + b_{1} x + b_{0}}

k

a_{i}, b_{i} \in R

a_{2 k} \neq 0, b_{2 k} \neq 0

b_{2 k} x^{2 k} + b_{2 k - 1} x^{2 k - 1} + . . . . . + b_{1} x + b_{0} = 0

f (x) = c o s (\frac{π}{x}), x \neq 0

k

(k + 1, 1), (2 k + 1, 3), (2 k + 2, 2 k)

Join BYJU'S Learning Program