Answer to Question #139619 in Molecular Physics | Thermodynamics for Sridhar

Question

Work done on heating one mole of monoatomic gas adiabatically through 20°C is W. Then the work done on heating 6 moles of rigid diatomic gas through the same change in temperature
Ans 10w

Answer 1

For the adiabatic process, from the first law of thermodynamics

$W = \Delta U$

The internal energy for one mole of ideal monoatomic gas is $U = \frac{3}{2}kT$ .

The internal energy for one mole of ideal diatomic gas is $U = \frac{5}{2}kT$ .

So, for the one mole of monoatomic gas $W_1 = \frac{3}{2} k \Delta T = 1.5 k \Delta T= W,$

For 6 moles of diatomic: $W_2 = 6 \cdot \frac{5}{2} k \Delta T = 15 k \Delta T$

$W_2 = 15 k \Delta T = 10 \cdot 1.5 k \Delta T = 10 W$.

Answer: 10W.