Answer:
1. K.E = 11.2239 kJ ≈ 11.224 kJ
2. [tex]C_{V} = 37.413 JK^{- 1}[/tex]
3. [tex]Q = 10.7749 kJ[/tex]
Solution:
Now, the kinetic energy of an ideal gas per mole is given by:
K.E = [tex]\frac{3}{2}mRT[/tex]
where
m = no. of moles = 3
R = Rydberg's constant = 8.314 J/mol.K
Temperature, T = 300 K
Therefore,
K.E = [tex]\frac{3}{2}\times 3\times 8.314\times 300[/tex]
K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ
Now,
The heat capacity at constant volume is:
[tex]C_{V} = \frac{3}{2}mR[/tex]
[tex]C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}[/tex]
Now,
Required heat transfer to raise the temperature by [tex]15^{\circ}[/tex] is:
[tex]Q = C_{V}\Delta T[/tex]
[tex]\Delta T = 15^{\circ} = 273 + 15 = 288 K[/tex]
[tex]Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ[/tex]
