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Ideal Gas Equation

An ideal mona...

Question

An ideal monatomic gas is at $P_{0}, V_{0}$ . It is taken to final volume $2 V_{0}$ when pressure is $P_{0} / 2$ in a process which is straight line on P-V diagram. Choose the correct statements.

Final temperature is equal to initial temperature.

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There is no change in internal energy.

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Work done by the gas is +

\frac{P_{0} V_{0}}{4}

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Heat is absorbed in the process.

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Solution

The correct options are
A Final temperature is equal to initial temperature.
B There is no change in internal energy.
D Heat is absorbed in the process.
(a) We have $P V = n R T$
If $T_{1}$ and $T_{2}$ are initial and final temperatures,
$P_{o} V_{o} = n R T_{1}$ and
$\frac{P_{o}}{2} 2 V_{o} = n R T_{2}$
$⟹ T_{1} = T_{2}$
(b) $T_{1} = T_{2} ⟹ Δ U = 0$
(c) Work done = area under the curve
$W = \frac{1}{2} \times P_{o} / 2 \times (2 V_{o} - V_{o}) + P_{o} / 2 \times (2 V_{o} - V_{o})$
$= \frac{1}{4} \times P_{o} \times V_{o} + \frac{1}{2} \times P_{o} \times V_{o}$
$= \frac{3}{4} \times P_{o} \times V_{o}$
(d) We have,
$Δ Q - Δ W = Δ U$
Since $Δ U = 0$ and $Δ W$ is positive (i.e.s work is done by the system),
$Δ Q > 0$ , heat is absorbed in the process.

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