Question

How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant volume? How much heat would need to be transferred to the system to raise the temperature by 15°C?

Answer 1

Answer:

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2. $C_{V} = 37.413 JK^{- 1}$

3. $Q = 10.7749 kJ$

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E = $\frac{3}{2}mRT$

where

m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

Temperature, T = 300 K

Therefore,

K.E = $\frac{3}{2}\times 3\times 8.314\times 300$

K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

Now,

The heat capacity at constant volume is:

$C_{V} = \frac{3}{2}mR$

$C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}$

Now,

Required heat transfer to raise the temperature by $15^{\circ}$ is:

$Q = C_{V}\Delta T$

$\Delta T = 15^{\circ} = 273 + 15 = 288 K$

$Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ$