At point A, the gas has volume 2.4 × 10-3 m3, pressure 1.6 × 105 Pa and temperature 300 K.

Question 9

(a) The first law of thermodynamics may be expressed in the form

ΔU = q + w.

(i) State, for a system, what is meant by:

1. +q

2. +w.

[2]

(ii) State what is represented by a negative value of ΔU. [1]

(b) An ideal gas, sealed in a container, undergoes the cycle of changes shown in Fig. 2.1.

Fig. 2.1

At point A, the gas has volume 2.4 × 10^-3 m³, pressure 1.6 × 10⁵ Pa and temperature 300 K.

The gas is compressed suddenly so that no thermal energy enters or leaves the gas during the compression. The amount of work done is 480 J so that, at point B, the gas has volume 8.7 × 10^-4 m³, pressure 6.6 × 10⁵ Pa and temperature 450 K.

The gas is now cooled at constant volume so that, between points B and C, 1100 J of thermal energy is transferred. At point C, the gas has pressure 1.6 × 10⁵ Pa and temperature 110 K.

Finally, the gas is returned to point A.

(i) State and explain the total change in internal energy of the gas for one complete

cycle ABCA. [2]

(ii) Calculate the external work done on the gas during the expansion from point C to

point A. [2]

(iii) Complete Fig. 2.2 for the changes from:

1. point A to point B

2. point B to point C

3. point C to point A.

Fig. 2.2

[4]

[Total: 11]

Reference: Past Exam Paper – June 2019 Paper 42 Q2

Solution:

(a)

(i)

1. energy transfer to the system by heating

2. (external) work done on the system

(ii) decrease in internal energy

(b)

(i) The initial and final temperatures are the same, hence the net change in internal energy around a full cycle is zero.

(ii)

work done = pΔV

work done = (–)1.6 × 10⁵ × (2.4 – 0.87)×10^-3

work done = (–)240 J

(iii)

{As given in the question,

From A to B,

the gas is compressed suddenly so that no thermal energy enters or leaves. (i.e. +q = 0)

the gas is compressed, so work is done ON the gas. +w = 480 J

ΔU = q + w = 0 + 480 = 480 J

From B to C,

the volume of the gas does not change (as seen in the graph), so ΔV = 0 and w = p ΔV = 0

the gas is cooled, so it loses thermal energy. q = - 1100 J

ΔU = q + w = -1100 + 0 = -1100 J

From C to A,

the gas expands, so it does work. w is negative. As calculated previously, w = - 240 J

in (b)(i), we have stated that the total change in internal energy in a closed cycle is zero.

let the change in internal energy from C to A be U.

(adding values in the last column should equal to 0)

U + 480 – 1100 = 0

U = 620 J

ΔU = q + w

620 = q – 240

q = 620 + 240 = 860 J}

A diffraction grating experiment is set up using yellow light of wavelength 600 nm.

Question 27

A diffraction grating experiment is set up using yellow light of wavelength 600 nm. The grating has a slit separation of 2.00 μm.

What is the angular separation (θ₂ – θ₁) between the first and second order maxima of the yellow light?

A 17.5°                        B 19.4°                        C 36.9°                        D 54.3°

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q27 & March 2019 Paper 12 Q29

Note: For March 2019 Paper 12 Q29: 'orange light' instead of 'yellow light'

Solution:

Answer: B.

For a diffraction grating: d sinθ = nλ

Wavelength λ = 600nm = 600×10^-9 m

Slit separation d = 2.00 μm = 2.0×10^-6 m

Angle θ_n = sin^-1 (nλ / d)

When n = 1, θ₁ = sin^-1 ({600×10^-9} / {2.0×10^-6}) = 17.458°

When n = 2, θ₂ = sin^-1 (2 × {600×10^-9} / {2.0×10^-6}) = 36.870°

Angular separation (θ₂ – θ₁) = 36.870 – 17.458 = 19.4°

A sinusoidal alternating voltage is to be rectified. (a) Suggest one advantage of full-wave rectification as compared with half-wave rectification.

Question 5

A sinusoidal alternating voltage is to be rectified.

(a) Suggest one advantage of full-wave rectification as compared with half-wave rectification. [1]

(b) The rectification is produced using the circuit of Fig. 7.1.

Fig. 7.1

All the diodes may be considered to be ideal.

The variation with time t of the alternating voltage applied to the circuit is shown in Fig. 7.2 and in Fig. 7.3.

(i) On the axes of Fig. 7.2, draw a graph to show the variation with time t of the potential difference across diode A. [1]

(ii) On the axes of Fig. 7.3, draw a graph to show the variation with time t of the potential difference across diode B. [1]

(c)

(i) On Fig. 7.1, draw the symbol for a capacitor, connected into the circuit so as to provide smoothing. [1]

(ii) Fig. 7.4 shows the variation with time t of the smoothed potential difference across the resistor R in Fig. 7.1.

1. State how the amount of smoothing may be increased. [1]

2. On Fig. 7.4, draw the variation with time t of the potential difference across resistor R for increased smoothing. [2]

Reference: Past Exam Paper – November 2009 Paper 41 Q7

Solution:

(a) Example:

More (output) power is available

There is less ripple for same smoothing capacitor

(any sensible suggestion)

(b)

(i) The curve should be drawn showing half-wave rectification

{Diode A conducts only when the upper terminal of the supply is positive. That is, when the line of the alternating voltage applied is positive, current flows and the p.d. across the resistor is positive. This means that the p.d. across the diode is ZERO. Here, we want to plot the graph of p.d. across the diode, not the resistor.}

(ii) The curve is similar to (i) but with a phase shift of 180°

{Diode B conducts only when the lower terminal of the supply is positive. That is, when the line of the alternating voltage applied is negative, current flows through the resistor and the p.d. across the resistor is positive. This means that the p.d. across the diode is ZERO. Again, here want to plot the graph of p.d. across the diode. So, the p.d. across the diode is positive when the applied alternating voltage is positive (that is, when the diode is reverse-biased).}

(c)

(i) A correct symbol, connected in parallel with R    

(ii)

1. By using a larger capacitor / second capacitor in parallel with R

(not increase R)

2. The curve has the same peak values, with a correct shape giving less ripple