One mole of an ideal gas undergoes a process p-p 01-V - V 0 2where p0 and V0 are constants- Find the temperature of the gas when V-V0 -

Given: #160;p=po1+(VVo)2Multiplying #160;both #160;sides #160;by #160;V, #160;we #160;getpV=poV1+(VVo)2pV=RT #160; #160; #160; #160; #16 ...

132

You visited us 132 times! Enjoying our articles? Unlock Full Access!

Abundance and Isotopes

One mole of a...

Question

One mole of an ideal gas undergoes a process
$p = \frac{p_{0}}{1 + (V / V_{0})^{2}}$
where p₀ and V₀ are constants. Find the temperature of the gas when V = V₀.

Open in App

Solution

Suggest Corrections

Similar questions

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 = P_{0} {(1 + {(\frac{2 V_{0}}{V})}^{2}]}^{- 1},

P_{0}, V_{0}

V = V_{0}

V = 2 V_{0}

\frac{n P_{o} V_{0}}{R}

n

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

P_{0}, V_{0}

V = V_{0}

V = 2 V_{0}

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

P_{0}

V_{0}

V = V_{0}

V = 2 V_{0}

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

P_{0}

V_{0}

V = V_{0}

V = 2 V_{0}

Join BYJU'S Learning Program