Question

[P+a÷v^2]V=RT, If P pressure ,T temperature,V Volume, R universal gas content .what are the dimensions of a?

Answer 1

Answer:

Given:

An equation has been provided as follows ;

$\bigg \{P + \dfrac{a}{ {V}^{2} } \bigg \}V = RT$

P is pressure , T is temperature , V is Volume , R is the universal Gas constant

To find:

Dimensions of "a"

Concept:

Only quantities of similar Dimensions can be added . For example ; P and a/(V²) can be added only because they have same Dimensions.

This means that a/(V²) has Dimensions as that of pressure.

Calculation:

As per our Concept, we can say that :

$\{P \} = \bigg \{\dfrac{a}{ {V}^{2} } \bigg \}$

$= > \bigg \{ \dfrac{force}{area} \bigg\} = \bigg \{\dfrac{a}{ {Volume}^{2} } \bigg \}$

$= > \bigg \{ \dfrac{ML {T}^{-2} }{ {L}^{2} } \bigg \} = \bigg \{ \dfrac{a}{ { ( {L}^{3}) }^{2} } \bigg \}$

$= > \{a \} = \bigg \{ M {L}^{5} {T}^{ - 2} \bigg \}$

So final answer is :

$\boxed{ \blue{ \sf{ \bold{ \huge{\{a \} = \bigg \{ M {L}^{5} {T}^{ - 2} \bigg \}}}}}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

Equation is : (P + a/v²)(v - b) = RT

_____________________________

To Find :

Dimensions of a

______________________________

Solution :

We have to find dimensions of a. So, we can terminate the equation (v - b)

So,

$\implies {\sf{P \: + \: \dfrac{a}{v^2} \: = \: RT}}$

As only same quantity can be added. So, dimensions of a/v² is same as P.

So, Pressure = a/v²

$\implies {\sf{\dfrac{a}{[L^3]^2} \: = \: [ML^{-1}T^{-2}]}} \\ \\ \implies {\sf{a \: = \: [ML^{-1}T^{-2}][L^3]^2}} \\ \\ \implies {\sf{a \: = \: [ML^{-1}T^{-2}][L^6]}} \\ \\ \implies {\sf{a \: = \: [ML^5T^{-2}]}}$