One mole of an ideal gas undergoes a process P p 0-1- V - V 02where P 0 and V 0 are constants- Find the temperature of the gas when V - V 0-

Here, P = P_0/1 + ( V/V_0 )^2 ⇒nRT/V = P_0/1+ ( V/V_0 )^2 [PV = nRT according to ideal gas equation] ⇒RT/V = P_0/1+ ( V/V_0 )^2[Since n= 1 mole] ⇒RT/V_0 = P ...

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Byju's Answer

Standard XII

Physics

Ideal Gas Equation from KTG

One mole of a...

Question

One mole of an ideal gas undergoes a process

$p \frac{p_{0}}{1 + (V / V_{0})^{2}}$

where $p_{0}$ and $V_{0}$ are constants. Find the temperature of the gas when $V = V_{0}$ .

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Solution

Here,

$P = \frac{P_{0}}{1 + {(\frac{V}{V_{0}})}^{2}}$

$\Rightarrow \frac{n R T}{V} = \frac{P_{0}}{1 + {(\frac{V}{V_{0}})}^{2}}$

[PV = nRT according to ideal gas equation]

$\Rightarrow \frac{R T}{V} = \frac{P_{0}}{1 + {(\frac{V}{V_{0}})}^{2}}$ [Since n= 1 mole]

$\Rightarrow \frac{R T}{V_{0}} = \frac{P_{0}}{1 + {(\frac{V}{V_{0}})}^{2}} [V = V_{0}]$

$\Rightarrow P_{0} V_{0} = R T (1 + 1)$

$\Rightarrow P_{0} V_{0} = R T \times 2$

T = $\frac{P_{0} V_{0}}{2 R}$

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$p \frac{p_{0}}{1 + (V / V_{0})^{2}}$

where $p_{0}$ and $V_{0}$ are constants. Find the temperature of the gas when $V = V_{0}$ .