• Physics 9702 Doubts | Help Page 190

Question 924: [Ideal Gas]

(a) State what is meant by an ideal gas.

(b) A storage cylinder for ideal gas has a volume of 3.0 × 10⁻⁴ m³. The gas is at a temperature of 23 °C and a pressure of 5.0 × 10⁷ Pa.

(i) Show that the amount of gas in the cylinder is 6.1 mol.

(ii) The gas leaks slowly from the cylinder so that, after a time of 35 days, the pressure has reduced by 0.40%. Temperature remains constant.

Calculate the average rate, in atoms per second, at which gas atoms escape from the cylinder.

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q3

Solution 924:

(a) An ideal gas is one which obeys the equation pV / T = constant

(b)

(i)

pV = nRT

{Temperature in Kelvin = 273 + 23 = 296 K}

5.0 × 10⁷ × 3.0 × 10^–4 = n × 8.31 × 296

Number of moles, n = 6.1 mol

(ii)

{pV = nRT. The pressure p is proportional to the amount of substance, n}

Pressure ∝ Amount of substance

{The final pressure is 100% - 0.40% = 99.6% of the original pressure. That is, 0.40% has been lost. So, the amount of substance loss is also 0.40% of the original amount since the pressure is proportional to the amount of substance. The original amount of substance (100%) is 6.1 mol, as calculated above.}

Loss = 0.40 / 100 × 6.1 mol = 0.0244 mol

{Since the average rate is asked in ‘atoms per second’, we need to convert the amount of moles into number of atoms.}

Loss = 0.0244 × 6.02 × 10²³ (atoms) = 1.47 × 10²² atoms

{Now, this amount of atoms has been lost in 35 days. To find the rate, we need to calculate the amount of atoms loss in one second.}

Rate = (1.47 × 10²²) / (35 × 24 × 60 × 60) = 4.9 × 10¹⁵ s^–1

Question 925: [Matter > Hooke’s law]

A number of similar springs, each having the same spring constant, are joined in four arrangements. Same load is applied to each.

Which arrangement gives the greatest extension?

Reference: Past Exam Paper – November 2008 Paper 1 Q21

Solution 925:

Answer: C.

Hooke’s law: F = ke

Extension, e = F / k

Since all the arrangements consist of more than 1 spring, we need to find the effective spring constant of the system in each arrangement.

For springs in series and in parallel, the following formulae for the effective spring constant apply:

In parallel: effective spring constant, k_eff = k₁ + k₂ + ….

In series: effective spring constant, 1/k_eff = 1/k₁ + 1/k₂ + ….

Therefore, considering a system of 2 springs, when springs are attached in parallel, the effective spring constant is greater and hence, the extension e (= F / k_eff) is smaller. Also, for springs attached in series, k_eff is smaller and hence the extension is larger.

Let the spring constant of each spring be k and let the load be F.

Arrangement A:

Effective spring constant, k_eff = k + k = 2k

Extension, e = F / k_eff = F / (2k) = 0.5 (F/k)

Arrangement B:

Effective spring constant, k_eff = [1/(k+k) + 1/(k+k)]^-1 = k

Extension, e = F / k_eff = F / k

Arrangement C:

Effective spring constant, k_eff = [1/(k+k) + 1/k]^-1 = 2k / 3

Extension, e = F / k_eff = F / (2k/3) = 3F / 2k = 1.5 (F/k)

Arrangement D:

Effective spring constant, k_eff = [1/(k+k+k) + 1/k]^-1 = 3k / 4

Extension, e = F / k_eff = F / (3k/4) = 4F / 3k = 1.33 (F/k)

Question 926: [Electromagnetism]

Two long straight vertical wires X and Y pass through a horizontal card, as shown in Fig.1.

The current in each wire is in the upward direction.

The top view of the card, seen by looking vertically downwards at the card, is shown in Fig.2.

(a) On Fig.2,

(i) draw four field lines to represent the pattern of the magnetic field around wire X due solely to the current in wire X,

(ii) draw an arrow to show the direction of the force on wire Y due to the magnetic field of wire X.

(b) The magnetic flux density B at a distance x from a long straight wire due to a current I in the wire is given by the expression

B = μ_oI / 2πx,

where μ_o is permeability of free space.

The current in wire X is 5.0 A and that in wire Y is 7.0 A. The separation of the wires is 2.5 cm.

(i) Calculate the force per unit length on wire Y due to the current in wire X.

(ii) Currents in the wires are not equal.

State and explain whether the forces on the two wires are equal in magnitude.

Reference: Past Exam Paper – November 2009 Paper 42 Q5

Solution 926:

(a)

(i) Field lines are concentric circles, in an anticlockwise direction. The separation of lines increases with the distance from the wire.

(ii) The direction is from Y towards X

(b)

(i)

(B = μ_oI / 2πx)

Flux density at wire Y = (4π ×10^-5) × 5.0 / (2 × π × 2.5×10^-2) = 4.0×10^-5T

Force per unit length = BI = (4.0×10^-5) × 7.0 = 2.8×10^-4N

(ii)

Either

The force depends on the products of the currents in the two wires. So, the forces on the two wires are equal.

Or

(Since this is an isolated system,) Newton’s third law applies. So, the forces on the two wires are equal.

Question 927: [Waves > Diffraction]

(a) Monochromatic light is diffracted by a diffraction grating. By reference to this, explain what is meant by

(i) diffraction,

(ii) coherence,

(iii) superposition.

(b) A parallel beam of red light of wavelength 630 nm is incident normally on a diffraction grating of 450 lines per millimetre.

Calculate the number of diffraction orders produced.

(c) The red light in (b) is replaced with blue light. State and explain the effect on the diffraction pattern.

Reference: Past Exam Paper – June 2012 Paper 23 Q6

Solution 927:

(a)

(i) Diffraction is the bending/spreading of light at an edge/slit. This occurs at each slit.

                               

(ii) Coherence is the constant phase difference between each of the waves.

(iii) (When the waves meet) the resultant displacement is the sum of the displacements of each wave

(b)

For diffraction grating: d sinθ = nλ

{The greatest possible value of θ for diffraction is 90°. So, to find the greatest order of diffraction, we take the angle as 90°. The value of sin (90°) is 1 and can thus be omitted in what follows.

d is the slit separation. There are 450 lines per millimeter (10^-3 m).

Separation of slits d = 10^-3 / 450 = 1 / (450 × 10³)      since 10^-3 = 1 / 10³}

Maximum order, n = d / λ = 1 / (450 × 10³ × 630 × 10^–9) = 3.52

{n can only be an integer so, n = 3}

Hence number of orders = 3

(c) The wavelength λ of blue light is less than the wavelength λ of red light. So more orders are seen. Each order is at a smaller angle than for the equivalent red.