# 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}m

^{3}. The gas is at a temperature of 23 °C and a pressure of 5.0 × 10

^{7}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 × 10

^{7}× 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 × 10

^{23}(atoms) = 1.47 × 10^{22}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 × 10

^{22}) / (35 × 24 × 60 × 60) = 4.9 × 10^{15}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}= k

_{1}+ k

_{2}+ ….

**In series:**effective spring constant, 1/k

_{eff}= 1/k

_{1}+ 1/k

_{2}+ ….

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 = Î¼

_{o}I / 2Ï€x,
where Î¼

_{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 = Î¼

_{o}I / 2Ï€x)
Flux density at wire Y = (4Ï€ ×10

^{-5}) × 5.0 / (2 × Ï€ × 2.5×10^{-2}) = 4.0×10^{-5}T
Force per unit length = BI = (4.0×10

^{-5}) × 7.0 = 2.8×10^{-4}N
(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 sinÎ¸ = nÎ»

{The greatest possible
value of Î¸ 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 × 10^{3}) since 10^{-3}= 1 / 10^{3}}
Maximum order, n = d / Î» = 1 / (450
× 10

^{3}× 630 × 10^{–9}) = 3.52
{n can only be an integer
so, n = 3}

Hence number of orders = 3

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

Please consider answering ALL of the following questions before October:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.3(c)(ii),Q.5(b),Q.9(b),Q.11(a)(ii)

04/O/N/08 Q.7(c)

04/M/J/09 Q.10(c)(ii),Q.11(b)(iii)

41/O/N/09 Q.6(a),(b)(i),Q.10

42/O/N/09 Q.7(b)(ii)

41/M/J/10 Q.6(a),Q.7(a)

51/M/J/10 Q.2(d)

42/O/N/10 Q.3(c)

41/M/J/11 Q.8(a)

For 04/M/J/09 Q.10(c)(ii), see solution 930 at

Deletehttp://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-191.html

in question 925 are the springs in series in option b since the formula (Effective spring constant, keff = (k + k) + (k + k)) says that?

ReplyDeleteSorry, there was a mistake above. I have corrected it.

DeleteThe 2 springs at the top is in parallel with each other and the 2 at the bottom are also in parallel with each other. The pair of string above is in series with the pair at the bottom.

Thanks