Temperature of one mole of helium gas is increased by 4-C- Thus- the increase in internal energy is-

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 difference will remain same) C_v=32R (a ...

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Standard XIII

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

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12 c a l

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3 c a l

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24 c a l

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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,
$△ U = \frac{3}{2} \times 2 \times 4 = 12 c a l$

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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: