Application of the first law of thermodynamics to isoprocesses Mark scheme

Mark Scheme

1.      B

2.      C

3.      D

4.      C

5.      B

6.      C

7.      Mean kinetic energy of atom µ absolute temperature [1]

 µ T or v² µ  [1]
Since the mass m of the atom is constant, we have: v µ  [1]
The temperature of 0 °C in kelvin is T = 273 K

The absolute temperature increases by a factor of  (= 36.6) [1]
Hence the speed will increase by a factor of  = 6.05 [1]
The speed of the atoms at 10 000 K = 1.3 × 6.05 » 7.9 km s^–1 [1]

8.      aThe particles have a range of speeds and travel in different directions.                           [1]

b i) Mean kinetic energy = [1]

                                           = 1.118 ´ 10^–19 J » 1.1 ´ 10^–19 J[1]

b ii)[1]

[1]

speed = 1.147 ´ 10⁴ m s^–1 » 11 km s^–1[1]

9.      (a) Realization that since pV = constant, the temperature must be the same i.e. 400 K / full calculation using gas law to get 400 K;                                                                       1

(b)         (i)      work done is area under curve;
and this is  =2400 J;                                                                                       2
Award [2] for correct bald answer.

(ii)     (Q = ∆U + W with) ∆U = 0;
so Q = 2400 J;                                                                                                         2
Award [0] for correct answer with no or wrong argument.

(c)     (i)      curve under given straight line starting at B and ending at A;                               1

(ii)     it would be less;
since the work done would be less / area under curve is
less (and ∆U = 0);                                                                                                   2

[8]