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Question

Answers

(a) $8.3145\times { 10 }^{ 3 }J{ K }^{ -1 }{ mol }^{ -1 }$

(b) $1.987\quad cal{ K }^{ -1 }{ mol }^{ -1 }$

(c) $0.083145\times { 10 }^{ 3 }{ dm }^{ 3 }bar{ K }^{ -1 }{ mol }^{ -1 }$

(d) $0.983145\quad { dm }^{ 3 }bar{ K }^{ -1 }{ mol }^{ -1 }$

Answer
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The molar gas constant is also called ideal gas constant and is denoted by the symbol R. Since the ideal gas equation is derived by using the Boyle’s law, Charles’ law, Gay-Lussac’s law and Avogadro law, the ideal gas constant is a combination of Boyle’s constant, Charles’ constant, Gay-Lussac’s constant and Avogadro’s constant.

Its value can be easily found out by multiplying the Boltzmann constant and the Avogadro number i.e.

$ R={ N }_{ A }{ k }_{ B }$

Where ${ N }_{ A }$ is the Avogadro number and ${ k }_{ B }$ is the Boltzmann constant.

Let us find out the dimensions of the gas constant using the ideal gas equation.

According to the ideal gas equation, the product of the number of moles of the gas with the temperature and the gas constant is equal to the product of the pressure and the volume:

PV=nRT

Therefore,

$\Rightarrow R=\cfrac { PV }{ nT } $

The pressure is force per unit area. Hence the dimensions will be:

$\Rightarrow R=\cfrac { V\times \cfrac { force }{ area } }{ amount\times temperature } $

Since, the area is square of the length and the volume is length raise to the power of 3, Hence:

$R=\cfrac { { length }^{ 3 }\times \cfrac { force }{ { length }^{ 2 } } }{ amount\times temperature } $

Therefore, $\Rightarrow R=\cfrac { { length }\times Force }{ amount\times temperature } $

Now, the product of force and the length is force, therefore:

$R=\cfrac { work }{ amount\times temperature } $

Using the above dimensional formula, we can find the SI unit of the gas constant. Since the SI unit of work is Joules, the SI unit for the amount of a substance is mole and the SI unit of the temperature is Kelvin, therefore:

$R=\cfrac { J }{ mol\times K } =J{ K }^{ -1 }{ mol }^{ -1 }$

Therefore the value of the gas constant in SI unit will be: $R=8.314\quad J{ K }^{ -1 }{ mol }^{ -1 }$

Now, 1 J= 4.128 cal, therefore

8.314 J will be= $4.128\quad cal\times \cfrac { 8.314\quad J }{ 1\quad J } =1.987\quad cal$