At constant pressure, a sample of helium gas has a volume of $800 m l$ at $27^{o} C$ . Calculate the temperature at which the volume will become $20$ % of its initial volume.

20 K

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

60^{o} C

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

150 K

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

- 213^{o} C

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

Open in App

Solution

The correct option is C $- 213^{o} C$
$V \propto T \frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}} ⟹ \frac{800 m l}{800 m l \times \frac{20}{100}} = \frac{300 K}{T_{2}} ⟹ T_{2} = 60 K = - 213 ℃$

Suggest Corrections

Similar questions

800 m l

27^{o} C

90^{0} C

27^{o} C

47^{o} C

27^{o} C

10 d m^{3}

d m^{3}

An ideal gas has volume $V_{0}$ at $27^{\circ} C$ .It is heated at constant pressure so that its volume becomes $2 V_{0}$ . The final temperature is

0.2

0^{o} C

273^{o} C

Join BYJU'S Learning Program