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Ideal Gas Equation

One mole of i...

Question

One mole of ideal gas undergoes a linear process as shown in figure. Its temperature expressed as a function of volume $V$ is

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

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T = \frac{P_{0} V}{R}

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T = \frac{P_{0} V}{R} (1 - \frac{V}{V_{0}})

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T = \frac{P_{0} V}{R} (1 - {(\frac{V}{V_{0}})}^{2})

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Solution

The correct option is D $T = \frac{P_{0} V}{R} (1 - \frac{V}{V_{0}})$
Comparing with $y = m x + c$ we have
$P = \frac{- P_{0}}{V_{0}} V + P_{0} . . . (1)$
Multiplying both sides with $V$
$P V = \frac{- P_{0}}{V_{0}} V^{2} + P_{0} V_{0} . . . (2)$

$P V = n R T . . . . . . . . (3)$
putting this expression in equation 2 and taking n=1 we get,
$T = \frac{P_{0} V}{R} (1 - \frac{V}{V_{0}})$

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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 $V = V_{0}$ .

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

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