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One Mole of an Ideal Gas Undergoes a Process P = P 0 1 + ( V V 0 ) 2 Where P 0 and V 0 Are Constants . Find the Temperature of the Gas When V = V 0 .

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प्रश्न

One mole of an ideal gas undergoes a process `P = (P_0)/(1+(V/V_0)^2` where `p_0` and `V_0` are constants . Find the temperature of the gas when `V=V_0` .

योग
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उत्तर

Given : 

`P = (p_0)/(1+(V/V_0)^2`

Multiplying both sides by V, we get

`pV = (p_0V)/(1+(V/V_0)^2)`

`pV = RT`              [From eq. (1)]        

Now , 

`RT = (p_0V)/(1+(V/V_0)^2)`

`T = 1/R ((p_0V_0)/(1+(V_0/V_0)^2))`       [`V=V_0`]

⇒ `T = (p_0V_0)/(2R)`

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अध्याय 24: Kinetic Theory of Gases - Exercises [पृष्ठ ३६]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 24 Kinetic Theory of Gases
Exercises | Q 38 | पृष्ठ ३६

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