You visited us 116 times! Enjoying our articles? Unlock Full Access!

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.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

There is no change in internal energy.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Work done by the gas is +

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Heat is absorbed in the process.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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.

Suggest Corrections

Similar questions

P_{0}

V_{0}

2 V_{0}

\frac{P_{0}}{2}

P - V

P_{0}

V_{0}

2 V_{0}

\frac{P_{0}}{2}

P - V

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}

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

P_{0}

V_{0}

V = V_{0}

V = 2 V_{0}

Join BYJU'S Learning Program