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 = × 6.02 × 1023 = 3.01 × 1023 molecules
As He is a monoatomic gas, degrees of freedom associated with each of its molecule is 3.
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 | = | |
Energy associated with 2g of He is | ||
U | = | (Total degrees of freedom) × |
U | = | f × |
U | = | 9.03 × 1023 × |
Energy at 27° c | = | 27 + 273 |
= | 300 k | |
U1 | = | 9.03 × 1023 × |
= | 1869.2 J | |
Energy at 127° c | = | 127 + 273 |
= | 400k | |
U2 | = | 9.03 × 1023 × |
= | 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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