At what temperature is kinetic energy of a gas molecule double that of its value of 27 degrees?

5. How many degrees of freedom are associated with 2g of He at NTP? Calculate the amount of heat energy required to raise the temperature of this amount from 27°c to 127°c.

Sol: Molecular weight of He = 4
∴Number of molecules in 4g of He = 6.02 × 1023 molecules
Number of molecules in 2g of He =

Now total degrees of freedom of 2g of He is
f	=	(Total number of molecules) × (Degrees of freedom per molecule)
f	=	3.01 × 1023 × 3
	=	9.03 × 1023
We know that energy associated with one degree of freedom per molecule
U	=	KT
Energy associated with 2g of He is
U	=	(Total degrees of freedom) × KT
U	=	f × KT
U	=	9.03 × 1023 × 1.38 × 10-23 × T
Energy at 27° c	=	27 + 273
	=	300 k
U1	=	9.03 × 1023 × 1.38 × 10-23 × 300
	=	1869.2 J
Energy at 127° c	=	127 + 273
	=	400k
U2	=	9.03 × 1023 × 1.38 × 1023 × 400
	=	2492.3 J
Heat energy required to raise the temperature from 27° c to 127° c is
U2 – U1	=	2492.3 – 1869.2
	=	623.1 J

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