| Date | May 2018 | Marks available | 3 | Reference code | 18M.2.HL.TZ1.2 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Determine | Question number | 2 | Adapted from | N/A |
Question
A closed box of fixed volume 0.15 m3 contains 3.0 mol of an ideal monatomic gas. The temperature of the gas is 290 K.
When the gas is supplied with 0.86 kJ of energy, its temperature increases by 23 K. The specific heat capacity of the gas is 3.1 kJ kg–1 K–1.
Determine, in kJ, the total kinetic energy of the particles of the gas.
Markscheme
ALTERNATIVE 1
average kinetic energy = \(\frac{3}{2}\)1.38 × 10–23 × 313 = 6.5 × 10–21 «J»
number of particles = 3.0 × 6.02 × 1023 = 1.8 × 1024
total kinetic energy = 1.8 × 1024 × 6.5 × 10–21 = 12 «kJ»
ALTERNATIVE 2
ideal gas so U = KE
KE = \(\frac{3}{2}\)8.31 × 131 × 3
total kinetic energy = 12 «kJ»
[3 marks]
Examiners report
[N/A]
