Question 18M.2.HL.TZ1.2
| Date | May 2018 | Marks available | [Maximum mark: 6] | Reference code | 18M.2.HL.TZ1.2 |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | Determine, Explain | Question number | 2 | Adapted from | N/A |
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.
[3]
ALTERNATIVE 1
average kinetic energy = 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 = 8.31 × 131 × 3
total kinetic energy = 12 «kJ»
[3 marks]
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.
Explain, with reference to the kinetic model of an ideal gas, how an increase in temperature of the gas leads to an increase in pressure.
[3]
larger temperature implies larger (average) speed/larger (average) KE of molecules/particles/atoms
increased force/momentum transferred to walls (per collision) / more frequent collisions with walls
increased force leads to increased pressure because P = F/A (as area remains constant)
Ignore any mention of PV = nRT.
[3 marks]