Temperature of one mole of helium gas is increased by 4- C- Thus- the increase in internal energy is-A- 3 calB- 24 calC- 12 calD- 6 cal

The correct option is C 12 cal nbsp;Heat capacity at constant volume is given as C_v= nbsp;( U T )_V U = C_V T using R=2 cal/K.mol T=4 ^oC or 4 K (as ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Heat Capacity

Temperature o...

Question

Temperature of one mole of helium gas is increased by $4^{\circ} C$ . Thus, the increase in internal energy is:

6 c a l

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

12 c a l

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

3 c a l

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

24 c a l

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

Open in App

Solution

The correct option is B $12 c a l$
Heat capacity at constant volume is given as
$C_{v} = {(\frac{△ U}{△ T})}_{V}$

$△ U = C_{V} △ T$
using $R = 2 c a l / K . m o l$
$△ T = 4^{o} C o r 4 K$ (as difference will remain same)
$C_{v} = \frac{3}{2} R$ (as $H e$ is monoatomic gas)
so,
$△ U = \frac{3}{2} \times 2 \times 4 = 12 c a l$

Suggest Corrections

Similar questions

Q. Temperature of one mole of helium gas is increased by

4^{\circ} C

. Thus, increase in internal energy is

Q. If

C_{v} = 4.96 c a l / m o l . K

then increase in internal energy when temperature of 2 moles of this gas is increased from

340 K

342 K

Q. Find the change in internal energy of

3 mol

of helium gas when its temperature is increased by

2 K

Q. Find the change in internal energy of

3 mol

of helium gas when its temperature is increased by

2 K

Q. To raise the temperature of 1 mole of a diatomic gas by

1^{\circ} C

at constant pressure, the required amount of heat is

120 c a l .

The amount of heat which goes as the internal energy is almost equal to:

Join BYJU'S Learning Program

Temperature of one mole of helium gas is increased by 4∘C. Thus, the increase in internal energy is:

The correct option is B 12 cal Heat capacity at constant volume is given as Cv=(△U△T)V △U=CV△T using R=2 cal/K.mol △T=4 oC or 4 K (as difference will remain same) Cv=32R (as He is monoatomic gas) so, △U=32×2×4=12 cal

Temperature of one mole of helium gas is increased by $4^{\circ} C$ . Thus, the increase in internal energy is: