A gas is found to obey the law $p^{2} V =$ constant, The initial temperature and volume are $T_{0}$ and $V_{0}$ . If the gas expands to a volume $3 V_{0}$ , its final temperature becomes:

\frac{T_{0}}{3}

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

\frac{T_{0}}{\sqrt{3}}

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

\sqrt{3} T_{0}

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

None of these

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

Open in App

Solution

The correct option is C $\sqrt{3} T_{0}$
Since, $p^{2} V =$ constant
${(\frac{R T}{V})}^{2} V =$ constant
$\Rightarrow \frac{T^{2}}{V} =$ constant
$\Rightarrow \frac{T_{2}}{T_{1}} = {(\frac{V_{2}}{V_{1}})}^{1 / 2}$
$\Rightarrow T_{2} = T_{0} \sqrt{3}$

Suggest Corrections

Similar questions

A gas found to obey the law $p^{2} V$ constant. The initial temperature and volume are $T_{0} a n d V_{0}$ . If the gas expands to a volume $2 V_{0}$ , its final temperature becomes

Q. A gas has found to obey

P^{2} V

= constant. The initial temperature and volume of gas are

T_{0}

V_{0}

. If the gas expands to volume

3 V_{0}

. Find the final temperature of gas

Q. A gas is found to obey the law

P^{2} V = c o n s t a n t

. The initial temperature and colume are

T_{0} a n d V_{0}

. If the gas expands to a volume

3 V_{0}

, then the final temperature becomes

Q. An ideal gas is found to obey an additional law

V P^{2}

= constant. The gas is initially at temperature T and volume V. When it expands to a volume 2V, the temperature becomes :

Q. Ten moles of a diatomic perfect gas are allowed to expand at constant pressure. The initial volume and temperature are

V_{0}

and

T_{0}

respectively. If

\frac{7}{2} R T_{0}

heat is transferred to the gas then the final volume and temperature are

Join BYJU'S Learning Program

A gas is found to obey the law p2V= constant, The initial temperature and volume are T0 and V0. If the gas expands to a volume 3V0, its final temperature becomes:

The correct option is C √3T0Since, p2V= constant(RTV)2V= constant⇒T2V= constant⇒T2T1=(V2V1)1/2⇒T2=T0√3

A gas is found to obey the law $p^{2} V =$ constant, The initial temperature and volume are $T_{0}$ and $V_{0}$ . If the gas expands to a volume $3 V_{0}$ , its final temperature becomes:

The correct option is C $\sqrt{3} T_{0}$
Since, $p^{2} V =$ constant
${(\frac{R T}{V})}^{2} V =$ constant
$\Rightarrow \frac{T^{2}}{V} =$ constant
$\Rightarrow \frac{T_{2}}{T_{1}} = {(\frac{V_{2}}{V_{1}})}^{1 / 2}$
$\Rightarrow T_{2} = T_{0} \sqrt{3}$