Velocity of sound in air at 0°C = 331 mls.
Audible frequencies, V₁ = 20 Hz, V₂ = 20 x 10³ Hz.
(i) Period (T):
T·tme penod \(T={1\over V}\), we get
\(∵\ T_1={1\over V_1}={1\over 20}=0.05s\)
\(T_2={1\over V_2}={1\over 20\times10^3}=0.055\times 10^{-3}s\)
∴ Time period range is 0.05 to 0.055 \(\times\)10^-3 s.
(ii) Wavelength ⋋:
From the formula \(\lambda ={v\over \lambda}\)
∴ \(\lambda_1={v\over V_1}={331\over 20}=16.55m\)
and
\(\lambda _2={v\over V_2}={331\over 20\times10^3}=0.0165m\)
∴ Wave length range is 16.55 m to 0.0165 m
(iii) Angular frequency ω:
From the formula, 0 = 27tV
∴ ω = 2πV₁= 2π \(\times\)20
= 40π rads^-1
and ω₂ = 2πV₂= 2π\(\times\)20\(\times\)10³
= 40π\(\times\)10³ rad/s.
∴  Angular frequency range is 40π to 40π\(\times\)10³ rad/s

Linear:
When a: solid rod of initial length I is heated through a temperature \(\triangle\)T, its final length (increased) is given by
L¹ = L + \(\triangle\)L = L (1 + \(\alpha\)\(\triangle\)T)
Where
\(\alpha\) is coefficient of linear expansion, It is given by
\(\alpha =\frac { \triangle L }{ L } \times \frac { 1 }{ \triangle T } \)
\(\alpha\) is defined as the increase in length per unit length per degree rise in temperature.
Superficial expansion:
When a solid, sheet of initial surface area A is heated through a temperature J). T, its final area (increased) is given by
A_1 ⁼ A + \(\triangle\)A = A (1 + \(\beta\)\(\triangle\)T)
\(\beta\) - coefficient of superficial expansion. It is given by
 \(\beta\) = \(\beta =\frac { \triangle A }{ A } \times \frac { 1 }{ \triangle T } \)
It is defined as the increase in surface area per unit area per degree rise in temperature.
Cubical expansion:
When a solid of initial volume V is heated through a temperature. J). T, its final volume is given by
V¹ = V + \(\triangle\)V = V (1 + \(\gamma \)\(\triangle\)T)
\(\gamma \) - coefficient of cubical expansion and it is defined as the increase in volume per unit volume per degree rise in temperature.
\(\gamma =\frac { \triangle V }{ V } \times \frac { 1 }{ \triangle T } \)

From ideal gas equation
pV = nRT (or) \(r=\frac{nRT}{V}\)
As gas expands such that
ρT² = a constant = C
So \((\frac{nRT}{V})T^{2}\) = C (or) T³ α V
(or) T³ = kV
differentiating it w.r.t.T, we have
\(3T^{2}=k \frac{dV}{dT}\)
Dividing it by T³=kV
\(\frac{3}{T}=\frac{1}{V}.\frac{dV}{dT}\)
Coefficient of volume expansion r = \(\frac{1}{V}.\frac{dV}{dT}=\frac{3}{T}\)

Efficiency of a Carnot engine: 
(i) Efficiency is defined as the ratio of work done by the working substance. in one cycle to the amount of heat extracted from the source. work done 
\(\eta =\frac { work \ done }{ Heat \ extracted } =\frac { W }{ { Q }_{ H } } \)
(ii) From the first law of thermodynamics, W=Q_H-Q_L
\(\therefore \quad \eta -\cfrac { { Q }_{ H }-{ Q }_{ L } }{ { Q }_{ H } } =1-\cfrac { { Q }_{ L } }{ { Q }_{ H } } \)
(iii)Applying isothermal conditions, we get,
Q_H = mRT_H In \(\left( { { V }_{ 2 } }/{ { V }_{ 4 } } \right) \)
\({ Q }_{ L }=\mu RT_{ L }\quad ln\left( { { V }_{ 3 } }/{ { V }_{ 4 } } \right) \) ...(3)
(iv) Here we omit the negative sign. Since we are interested in only the amount of heat (Q_L) ejected into the sink, we have
\(\cfrac { { Q }_{ L } }{ { Q }_{ H } } =\frac { { T }_{ L }ln\left( { { V }_{ 3 } }/{ { V }_{ 4 } } \right) }{ { T }_{ H }ln\left( { { V }_{ 2 } }/{ { V }_{ 1 } } \right) } \)        ...(4)
By applying adiabatic conditions, we get,
\({ T }_{ H }{ V }_{ 2 }^{ \gamma -1 }={ T }_{ L }{ V }_{ 3 }^{ \gamma -1 }\)
\({ T }_{ H }{ V }_{ 1 }^{ \gamma -1 }={ T }_{ L }{ V }_{ 4 }^{ \gamma -1 }\)
By dividing the above two 'equations, we get         
\(\left( \cfrac { { V }_{ 2 } }{ { V }_{ 1 } } \right) ^{ \gamma -1 }=\left( \cfrac { { V }_{ 3 } }{ { V }_{ 4 } } \right) ^{ \gamma -1 }\)
Which implies that  \(\cfrac { { V }_{ 2 } }{ { V }_{ 1 } } =\cfrac { { V }_{ 3 } }{ { V }_{ 4 } } \)   ...(5)
Substituting equation (5) in (4), we get
\(\cfrac { { Q }_{ L } }{ { Q }_{ H } } =\cfrac { { T }_{ L } }{ { T }_{ H } } \)
\(\therefore \)The efficiency \(\eta =1-\cfrac { { T }_{ L } }{ { T }_{ H } } \)
Note: TL and TH should be expressed in Kelvin, scale.

James Prescott Joule showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were attached with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity both the masses lose potential energy equal to 2 mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel. This causes
 
a rise in temperature of the water. This implies that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat. He found that to raise 1 g of an object by 1°C, 4.186 J of energy is required. In earlier days the heat was measured in calorie.
1 cal = 4.186 J
This is called Joule's mechanical equivalent of heat.

From the principle of moments,
\(100 \times g \times x_{1}=m\times g \times x_{2}\)
\(100 \times cos 53^{0}=m\times cos 37^{0}\) .................(1)

Where, x₁ and x₂ are the arm lengths.
The right angle triangle with angles \(37^{0}, 53 ^{0} \ and \ 90^{0}\) is a special triangle which has the respective sides in the ratio, 3:4:5 as shown in the diagram.
Substituting the values in equation (1),

\(100 \times cos 53^{0}=3\times cos 37^{0}\)
\(100 \times \frac{3}{5}=m\times \frac{4}{5}\)
\(m=100\times \frac{3}{4}\)
m = 75 kg

For both persons, the Kinematic equations are the same, with u=0,a_c =g and a_moon  =\(g\over 6\)
a_e = g and a_m =\(g\over 6\)
For a person on earth, V_earth = \(\sqrt{2gh}=\sqrt{2\times 10\times 100}\)
Hence, V_earth =\(\sqrt{2000} ms^{-1}\) gives the velocity at the ground, on earth.
Similarly, for a person on the moon,
 V_earth =\(\sqrt{2gh\over6}={\sqrt{2000}\over \sqrt{6}}ms^{-1}\)
The person on earth reaches ground with greater velocity than the person on the moon.

Galileo's experiment with. the second plane (a) at same inclination angle Cisthe first (b) with increased smoothness (c) with reduced angle of inclination (d) with zero angle of inclination
When a ball rolls from the top of an inclined plane to its bottom, after reaching the ground it moves some distance and continues to move on to another inclined plane of same angle of inclination as shown in the Figure (a). By increasing the smoothness of both the inclined planes, the ball reach almost the same height (h) from where it was released (L1) in the second plane (L2) [figure (b)]. The motion of the ball is then observed by varying the angle of inclination of the second plane keeping the same smoothness. If the angle of inclination is reduced, the ball travels longer distance in the second plane to reach the same height [figure (c)). When the angle of inclination is made zero, the ball moves forever in the horizontal direction [figure (d)]. If the Aristotelian idea were true, the ball would not have moved in the second plane even if its smoothness is made maximum since no force acted on it in the horizontal direction.

Acceleration of a body 'a' = 25 ms^-2
Distance travelled during n^th Second
\(=u+\frac { a }{ 2 } (2n-1)\)
\(\because\) Distance travelled during 20^th Second
\(=u+\frac { a }{ 2 } (2n-1)\)
Initial velocity u =0, a = 25 ms^-2
Distance travelled = 0 +\(\frac { 25 }{ 2 } (2\times 20-1)\)
\(=\frac { 25 }{ 2 } (40-1)\)
\(=\frac { 25 }{ 2 } (139)\)
= 487.5 metre.

M.I. of a H₂ molecule about an axis passing through Center of Mass.
I = 2m.\(\frac{a^{2}}{4}\)
\(=\frac { 2\times 1.7\times { 10 }^{ -27 }\times { (4\times { 10 }^{ -10 }) }^{ 2 } }{ 4 } \)
\(=\frac { (3.4\times { 10 }^{ -27 })\times (16\times { 10 }^{ -20 }) }{ 4 } \)
= 13.6 \(\times\)10^-47 (or) 1.36\(\times\)10^-46 kg m²

Mass of a body(m) =\(\frac{500}{1000}\)kg = 0.5 kg
Horizontal Force (F) = 1N
Time(t) = 20s.
Therefore, acceleration of a body.
a = F/m =\(\frac{1}{0.5}=2ms^{-2}\)
Distance travelled s = ut+\(\frac{1}{2}+at^{2}\)
S =0 x 20 +\(\frac{1}{2}\times2\times(20)^{2}\)
= 0+\(\frac{1}{2}\times400\times2\)
Displacement s¹= 400 m
Workdone = F\(\times\)s = 1\(\times\)400
\(\boxed{=400}\) J; v = u+at = 0+2\(\times\)20
v = 40
Change in k.E =\(\frac{1}{2}m(v^{2}-u^{2})\)
=\(\frac{1}{2}\times0.5\times(40^{2}-0)\)
=\(=\frac{1}{2}\times 0.5\times 1600\)
\(\boxed{=400\ J}\)

Propagation of errors
A number of measured quantities may be involved in the final calculation of an experiment. Different types of instruments might have been used for taking readings. Then we may have to look at the errors in measuring various quantities, collectively.
The error in the final result depends on
(i) The errors in the individual measurements
(ii) On the nature of mathematical operations performed to get the final result. So we should know the rules to combine the errors.
The various possibilities of the propagation or combination of errors in different mathematical operations are discussed below:
(i) Error in the sum of two quantities:
Let A\(\triangle\) and \(\triangle\)B be the absolute errors in the two quantities A and B respectively. Then, 
Measured value of A = A \(\pm\triangle\) A
Measured value of B = B \(\pm\triangle\) B
Consider the sum, Z = A + B
The error \(\triangle\) Z in Z is the given by
Z  \(\pm\triangle\) Z = (A \(\pm\triangle\)A) + ( B \(\pm\triangle\) B)
= ( A + B ) \(\pm\) (\(\triangle\)A+ \(\triangle\) B)
= Z \(\pm\) ( \(\triangle\) A + \(\triangle\) B )
(or) \(\triangle\)Z = \(\triangle\) A+ \(\triangle\) B
The maximum possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities.
(ii) Error in the difference of two quantities:
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities, A and B, respectively. Consider the product Z = AB
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities, A and B, respectively. Consider the product Z = AB
The error \(\triangle\)Z in Z is given by \(Z \pm \Delta Z=(A \pm \Delta A)(B \pm \Delta B)\)
\(=(A B) \pm(A \Delta) \pm(B \Delta A) \pm(\Delta A . \Delta B)\)
Dividing L.H.S by Z and R.H.S by AB, we get,
\(1 \pm \frac{\Delta Z}{Z} \cdot 1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm \frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\)
As \(\triangle\)A/A, \(\triangle\)B/B are both small quantities, their product term \(\frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\) can be neglected. The maximum fractional error in Z is
\(\frac{\Delta Z}{Z}=\pm\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right)\)
The maximum error in difference of two quantities is equal to the sum of the absolute errors in the individual quantities.
(iii) Error in the division or quotient of two quantities
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities A and B respectively.
Consider the quotient, \(\mathrm{Z}=\frac{A}{B}\)
The error \(\triangle\)Z in Z is given by
\(Z \pm \Delta Z =\frac{A \pm \Delta A}{B \pm \Delta B}=\frac{A\left(1 \pm \frac{\Delta A}{A}\right)}{B\left(1 \pm \frac{\Delta B}{B}\right)} \)
\(=\frac{A}{B}\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)^{-1}\)
or \(Z \pm \Delta Z=Z\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)\)
[using (1+x) n \(\approx\) 1+n x, when x<1]
Dividing both sides by Z, we get,
\(1 \pm \frac{\Delta Z}{Z} =\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)  \)
\(=1 \pm \frac{\Delta Z}{Z} \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\)
As the terms \(\triangle\)A /A and \(\triangle\)B/ B are small, their product term can be neglected. The maximum fractional error in Z is given by
\(\frac{\Delta Z}{Z}=\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right)\)
The maximum fractional error in the quotient of two quantities is equal to the sum of their individual fractional errors.

We can use Kepler’s third law,
T² c (R_E + h)³
T^2/3 = c^1/3(R_E + h)
\({ \left( \frac { { T }^{ 2 } }{ c } \right) }^{ 1/3 }\)(R_E + h)
\({ \left( \frac { { T }^{ 2 }{ GM }_{ E } }{ 4\pi ^{ 2 } } \right) }^{ \frac { 1 }{ 3 } }=\left( { R }_{ E }+h \right) ;\)
\(c=\frac { { 4\pi }^{ 2 } }{ { GM }_{ E } } \)
\(h={ \left( \frac { { T }^{ 2 }{ GM }_{ E } }{ 4\pi ^{ 2 } } \right) }^{ 1/3 }-{ R }_{ E }\)
Here h is the distance of the Moon from the surface of the Earth. Here,
R_E - radius of the Earth = 6.4\(\times\)10⁶ m
M_E - mass of the Earth = 6.02\(\times\)10²⁴ kg
G – Universal gravitational
constant = 6.67\(\times\)10⁻¹¹ \(\frac { { Nm }^{ 2 } }{ { kg }^{ 2 } } \) 
By substituting these values, the distance to the Moon from the surface of the Earth is calculated to be 3.77\(\times\)10⁵ km

The gravitational potential energy U(r) can be calculated by finding the sum of gravitational potential energy of each pair of particles.
\(U=-\frac { { Gm }_{ 1 }m_{ 2 } }{ { r }_{ 12 } } -\frac { { Gm }_{ 1 }m_{ 3 } }{ { r }_{ 13 } } -\frac { { Gm }_{ 1 }m_{ 4 } }{ { r }_{ 14 } } -\frac { { Gm }_{ 2 }m_{ 3 } }{ { r }_{ 23 } } -\frac { { Gm }_{ 2 }m_{ 4 } }{ { r }_{ 24 } } -\frac { { Gm }_{ 3 }m_{ 4 } }{ { r }_{ 34 } } \)
Here r₁₂, r₁₃ ... are distance between pair of particles
\({ r }_{ 14 }^{ 2 }={ R }^{ 2 }+{ R }^{ 2 }=2{ R }^{ 2 }\)
\({ r }_{ 14 }=\sqrt { 2 } R={ r }_{ 12 }={ r }_{ 23 }={ r }_{ 34 }\)
r₁₃ = r =₂₄ = 2R 
\(U=-\frac { { Gm }_{ 1 }m_{ 2 } }{ \sqrt { 2 } R } -\frac { { Gm }_{ 1 }m_{ 3 } }{ { 2R } } -\frac { { Gm }_{ 1 }m_{ 4 } }{ { \sqrt { 2 } R } } -\frac { { Gm }_{ 2 }m_{ 3 } }{ \sqrt { 2 } R } -\frac { { Gm }_{ 2 }m_{ 4 } }{ 2R } -\frac { { Gm }_{ 3 }m_{ 4 } }{ \sqrt { 2 } R } \)
\(U=-\frac { G }{ R } \left[ \frac { { Gm }_{ 1 }m_{ 2 } }{ \sqrt { 2 } } -\frac { { Gm }_{ 1 }m_{ 3 } }{ { 2 } } -\frac { { Gm }_{ 1 }m_{ 4 } }{ { \sqrt { 2 } } } -\frac { { Gm }_{ 2 }m_{ 3 } }{ \sqrt { 2 } } -\frac { { Gm }_{ 2 }m_{ 4 } }{ 2 } -\frac { { Gm }_{ 3 }m_{ 4 } }{ \sqrt { 2 } } \right] \)
If all the masses are equal, then m₁ = m₂ = m₃ = m₄ = M
\(U=-\frac { G{ M }^{ 2 } }{ R } \left[ \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ { 2 } } +\frac { 1 }{ { \sqrt { 2 } } } +\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } +\frac { 1 }{ \sqrt { 2 } } \right] \)
\(U=-\frac { G{ M }^{ 2 } }{ R } \left[ 1+\frac { 4 }{ \sqrt { 2 } } \right] \)
\(U=-\frac { G{ M }^{ 2 } }{ R } \left[ 1+2\sqrt { 2 } \right] \)
The gravitational potential V(r) at a point O is equal to the sum of the gravitational potentials due to individual mass. Since potential is a scalar, the net potential at point O is the algebraic sum of potentials due to each mass.
\({ V }_{ o }(r)\frac { { Gm }_{ 1 } }{ R } -\frac { { Gm }_{ 2 } }{ { R } } -\frac { { Gm }_{ 3 } }{ { R } } -\frac { { Gm }_{ 4 } }{ R } \)
If m₁ = m₂ = m₃ = m₄ = M 
\({ V }_{ o }(r)\frac { { 4GM } }{ R } \)

The distance between OA and OB are the same and hence, the waves starting from A and B reach O after covering equal distances (equal path lengths). Thus, the path difference between two waves at O is zero.
OA − OB = 0
Since the waves are in the same phase, at the point O, the phase difference between two waves is also zero. Thus, the resultant intensity at the point O is maximum.
Consider a point Y, such that the path difference between two waves is λ. Then the phase difference at Y is \(\Delta \varphi=\frac{2 \pi}{\lambda} \times \Delta r=\frac{2 \pi}{\lambda} \times \lambda=2 \pi\)
Therefore, at the point Y, the two waves from A and B are in phase, hence, the intensity will be maximum.
Consider a point X, and let the path difference the between two waves be \(\frac { \lambda }{ 2 } \) Then the phase difference at X is
\(\Delta \varphi =\frac { 2\pi }{ \lambda } \frac { \lambda }{ 2 } \pi \)
Therefore, at the point X, the waves meet and are in out of phase, Hence, due to destructive interference, the intensity will be minimum.

a. Since k₁ and k₂ are parallel, k_u = k₁ + k₂ Similarly, k₃ and k₄ are parallel, therefore, k_d = k₃ + k₄ But k_u and k_d are in series,
therefore, \({ k }_{ eq }=\frac { { k }_{ u }{ k }_{ d } }{ { k }_{ u }+{ k }_{ d } } \)
If all the spring constants are equal then, k₁ = k₂ = k₃ = k₄ = k
Which means, k_u = 2k and k_d = 2k
Hence, \({ k }_{ eq }=\frac { { 4k }^{ 2 } }{ 4k } =k\)
b. Since k₁ and k₂ are parallel, k_A = k₁ + k₂ Similarly, k₄ and k₅ are parallel,
therefore, k_B = k₄ + k₅
But k_A, k₃, k_B, and k₆ are in series,
therefore, \(\frac { 1 }{ { k }_{ eq } } =\frac { 1 }{ { K }_{ A } } +\frac { 1 }{ { K }_{ 3 } } +\frac { 1 }{ { K }_{ B } } +\frac { 1 }{ { K }_{ 6 } } \)
If all the spring constants are equal
then, k₁ = k₂ = k₃ = k₄ = k₅ = k₆ = k
which means, k_A = 2k and k_B = 2k
\(\frac { 1 }{ { k }_{ eq } } =\frac { 1 }{ { 2K } } +\frac { 1 }{ { K } } +\frac { 1 }{ { 2K } } +\frac { 1 }{ { K } } =\frac { 3 }{ { K } } \)
\({ k }_{ eq }=\frac { k }{ 3 } \)

To draw 25 L of water from the well, the student has to do work against gravity by burning his energy.
Mass of the water = 25 L = 25 kg (1L = 1kg )
The work required to draw 25 kg of water = gravitational potential energy gained by water.
W = mgh = 25\(\times\)10\(\times\)25 = 6250 J
The total energy gained from the food = 200 food cal = 200 kcal.
= 200\(\times\)10³\(\times\)4.186 J = 8.37\(\times\)10⁵ J
If we assume that by using this energy the student can drawn ‘n’ pots of water from the well, the total energy spent by him = 8.37\(\times\)10⁵ J = nmgh
\(n=\frac{8.37 \times 10^{5} \mathrm{~J}}{6250 \mathrm{~J}} \approx 134\)
This n is also equal to the number of trees that he can water. Is it possible to draw 134 pots of water from the well just by having breakfast? No. Actually the human body does not convert entire food energy into work. It is only approximately 20% efficient. It implies that only 20% of 200 food calories is used to draw water from the well. So 20% of the 134 is only 26 pots of water. It is quite meaningful. So he can water only 26 trees. The remaining energy is used for blood circulation and other functions of the body. It is to be noted that some energy is always ‘wasted’. 

By following the law of triangular addition, the resultant vector is given by \(\vec R\) = \(\vec A\) + \(\vec B\) as illustrated below.
The magnitude of the resultant vector \(\vec R\) is given by
\(R=|\vec R|=\sqrt{5^2+7^2+2\times 5\times 7\cos 60^o}\)
\(R=\sqrt{25+49+\frac{70\times 1}{2}}=\sqrt{109}\) units

The angle \(\alpha\) between \(\vec R\) and \(\vec A\) is given by
\(\tan\alpha=\frac{B\sin\theta}{A+B\cos\theta}\)
\(\tan\alpha=\frac{7\times\sin60^o}{5+7\cos60^o}=\frac{7\sqrt{3}}{10+7}=\frac{7\sqrt{3}}{17}\) = 0.713
\(\therefore\alpha=35^o\)

The average size of a class is 6m length, 5 m breadth and 4 m height. The volume of the room V = 6\(\times\)5 \(\times\)4 = 120m³. We can determine the number of mole. At room temperature 300K, the volume of a gas occupied by any gas is equal to 24.6L.
The number of mole \(\mu =\frac { 120{ m }^{ 2 } }{ 24.6\times { 10 }^{ -3 }{ m }^{ 3 } } \approx 4878mol\)
Air is the mixture of about 20% oxygen, 79% nitrogen and remaining one percent are argon, hydrogen, helium, and xenon. The molar mass of air is 29 g mol^-1.
So the total mass of air in the room m = 4878\(\times\)29 = 141.4kg

We can take air molecules in the tire as an ideal gas. The number of molecules and the volume of tire remain constant. So the air molecules at 27°C satisfies the ideal gas equation P₁V₁ = NkT₁ and at 39°C it satisfies P₂V₂ = NkT₂
But we know
V₁ = V₂ = V
\(\frac { { P }_{ 1 }V }{ { P }_{ 2 }V } =\frac { Nk{ T }_{ 1 } }{ Nk{ T }_{ 2 } } \)
\(\frac { { P }_{ 1 } }{ { P }_{ 2 } } =\frac { { T }_{ 1 } }{ { T }_{ 2 } } \)
\( { P }_{ 2 } =\frac { { T }_{ 1 } }{ { T }_{ 2 } } { P }_{ 1 }\)
\({ P }_{ 2 }=\frac { 312K }{ 300K }\times 240\times{ \ 10 }^{ 3 }Pa=249.6Pa\qquad \)