Question

How many degrees of freedom are associated with 2 grams of helium at NTP? Calculate the amount of heat energy required to raise the temp. of this amount from 27°C to 127°C. Given Boltzmann constant, kB = 1.38 × 10^-16 erg molecule^-1 K^-1 and Avogadro's number = 6.02 × 10^23.​

Answer 1

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As , He is monotomic, each molecule has three degree of freedom. U=U2-U1=2492.3-1869.2=623.1joule

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Answer 2

Given :

• Mass of helium = 2 gram
• Botzmann constant, $\sf k_B = 1.38 \times 10^{-16}\: erg \:molecule^{-1} K^{-1}$
• Avogadro's number = 6.02 × 10²³

To Find :

The amount of heat energy required to raise the temp. of this amount from 27°C to 127°C = ?

Solution :

We have :

• Mass of helium = 2 gram

We know that, Atomic weight of helium = 4

So, number of atoms in 2 gram = $\sf \dfrac{6.02 \times 10^{23} \times 2}{4} = 3.01 \times 10^{23}$

As, Helium is monoatomic, each molecule has three degrees of freedom.

Hence, Total no. of degrees of freedom = 3 × 3.01 × 10²³ = 9.03 × 10²³

As, energy associated with one degree of freedom per molecule = $\sf \dfrac{1}{2}k_B T$

Hence, At T₁ = 27°C (300K), heat content of gas :

$\sf U_1 = 9.03 \times 10^{23} \times \dfrac{1}{2} k_B T_1 \\ \sf =9.03 \times 10^{23} \times \dfrac{1}{2} \times 1.38 \times 10^{ - 23} \times 300J \\ \sf = 1869.2J$

Similarly, At T₂ = 127°C (400K), heat content of gas :

$\sf U_2 = 9.03 \times 10^{23} \times \dfrac{1}{2} k_B T_1 \\ \sf =9.03 \times 10^{23} \times \dfrac{1}{2} \times 1.38 \times 10^{ - 23} \times 400J \\ \sf = 2492.3J$

Hence, energy required to raise the temp. of gas from 27°C to 127°C is :

$\sf U = U_2 - U_1 = 2492.3J - 1869.2J = 623.1J$

$\pink{ \underline{\bf Hence, \: Answer \: is \: 623.1 J}}$