How many degrees of freedom are associated with 2 grams of He at NTP ?

Question

How many degrees of freedom are associated with 2 grams of He at NTP ? (A) 3 (B) 3.01 × 1023 (C) 9.03 × 1023 (D) 6

Tardigrade Admin · Accepted Answer

Moles of He =(2/4)=(1/2) Molecules =(1/2) 6.02 1023=3.01 × 1023 As there are 3 degrees of freedom corresponding to 1 molecule of a monatomic gas Total degrees of freedom =3 × 3.01 × 1023

Q. How many degrees of freedom are associated with $2$ grams of $He$ at NTP ?

$3$

$3.01 \times 1 0^{23}$

$9.03 \times 1 0^{23}$

$6$

Moles of $He = \frac{2}{4} = \frac{1}{2}$
Molecules $= \frac{1}{2} 6.02 1 0^{23} = 3.01 \times 1 0^{23}$
As there are 3 degrees of freedom corresponding to 1 molecule of a monatomic gas
Total degrees of freedom $= 3 \times 3.01 \times 1 0^{23}$

Q. How many degrees of freedom are associated with 2 grams of He at NTP ?

3

3.01×1023

9.03×1023

6

Moles of He=42​=21​ Molecules =21​6.021023=3.01×1023 As there are 3 degrees of freedom corresponding to 1 molecule of a monatomic gas Total degrees of freedom =3×3.01×1023

Q. How many degrees of freedom are associated with $2$ grams of $He$ at NTP ?

$3$

$3.01 \times 1 0^{23}$

$9.03 \times 1 0^{23}$

$6$

Moles of $He = \frac{2}{4} = \frac{1}{2}$
Molecules $= \frac{1}{2} 6.02 1 0^{23} = 3.01 \times 1 0^{23}$
As there are 3 degrees of freedom corresponding to 1 molecule of a monatomic gas
Total degrees of freedom $= 3 \times 3.01 \times 1 0^{23}$