3 Marks Question - Page 2 - Physics STD 11 Science Questions - Vidyadip

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3 Marks Question

Question 513 Marks

200J of work is done on a gas to reduce its volume by compressing it. If this change is done under adiabatic conditions, find out the change in internal energy of the gas and also the amount of heat absorbed by the gas.

Answer

In adiabatic changes, dQ = 0
$\therefore\text{dQ}=\text{dU}+\text{dW}=0$
$\text{dU}=-\text{dW}=-(-200\text{J})=200\text{J}$
Internal energy increases by 200J. Heat absorbed is zero.

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Question 523 Marks

Consider a Carnot’s cycle operating between T₁ = 500K and T₂=300K producing 1k J of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

Answer

Efficiency of Carnot's engine $\mu=1-\frac{\text{T}_{2}}{\text{T}_{1}}$
Tempreature of source or reservior = T₁ = 500K
Tempreature of sink T₂ = 300K
$\therefore\mu=1-\frac{\text{T}_{2}}{\text{T}_{1}}$
$\frac{\text{Output work}}{\text{Input work}(E)}=1-\frac{300}{500}$
$\frac{1000\text{J}}{\text{x}}=1-0.6$
$\text{x}=\frac{1000}{\text{x}}=0.4$
$\text{x}=\frac{1000}{0.4}=2500 \text{J} .$

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Question 533 Marks

What are isotherms? Explain the process of liquiefication using it.

Answer

Isotherms are P-V graphs drawn at same temperature. When pressure is increased, volume is reduced at constant temperature. Beyond certain level, the pressure remains same for certain reduction in volume and then the pressure increases sharply. As temperature is increased, the constant portion reduces and at a temperature called critical temperature, the flat portion is absent. In each isotherm, before the flat portion is reached, the substance is in gaseous state and beyond the flat portion, it is in the liquid state. During the flat portion of P-V graph, the gas is under liquefication and so both gaseous and liquid states co-exist. Beyond T. it is not possible to liquefy the gas whatever large the pressure may be.

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Question 543 Marks

During India-Pakistan war, a soldier discovered that his lead bullet just melted when stopped by an obstacle. Calculate the velocity of the bullet if its temp. was 47.6°C. Given: melting point of lead = 327°C. Specific heat of lead = 0.03 cal g^-1°C^-1, latent heat of fusion of lead = 6 cal g^-1 and J = 4.2 × 10⁷ erg car^-1.
Assume that no heat is lost.

Answer

Increase in temperature,
$\theta=(327-47.6)^\circ\text{C}=279.4^\circ\text{C}$
Let m be the mass of the bullet.
Heat required,
$\text{Q}=\text{mS}\theta+\text{mL}$
or $\text{Q}=\text{m}(\text{S}\theta+\text{L})$
$=\text{m}(0.03\times279.4+6)$
$=14.38\text{m cal}$
Work done, $\text{W}=\frac12\text{mv}^2$
Where v is the velocity of bullet,
Now, $\text{W}=\text{JQ}$
$\therefore\frac12\text{mv}^2=4.2\times10^7\times14.38\text{m}$
$\text{v}^2=2\times4.2\times10^7\times14.38$
$\text{v}=3.48\times10^4\text{cm s}^{-1}$

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Question 553 Marks

What is the Internal energy in the process of vapourisation?

Answer

During vapourisation, volume increases. So work done = P(V_f - V_i) and temperature does not change. So, dQ = mL_v
$\therefore$ from first law of thermodynamics, dU= dQ - DW
= mL_v - P(V_f - V_i).

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Question 563 Marks

Two thermometers are constructed in the same way except that one has a spherical bulb and the other has an elongated cylindrical bulb. Which one will respond quickly to the temperature changes? Why?

Answer

Thermometer with the cylindrical bulb. It is because the area of a cylinderical surface is greater than a spherical surface. Therefore, mercury in the cylindrical bulb reaches the temperature of the surrounding earlier.

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Question 573 Marks

State law of equi-partition of energy. Use this law to calculate specific heats of monoatomic, diatomic and triatomic gases.

Answer

According to the equi partition of energy, each degree of freedom will contribute an equal energy of $\frac12\text{R}$ per mole. In mono, di and triatomic gases hawing 3, 5 and 6 or 7 degrees of freedom, the internal energy will be $\frac32\text{R},\frac52\text{RT}$ and $\frac62\text{RT}$ or $\frac72\text{RT}.$
Using dU = nC_VdT for mole, we get $\text{C}_\text{V}=\frac32\text{R},\frac{5}{2}\text{R},$ and 3R or $\frac72\text{R}$ for the three gases respectively.

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Question 583 Marks

What are 'Super heated water' and 'Super cooled vapour'?

Answer

Water in liquid phase at a temperature above 100°C and a pressure more than 1atm is called as super heated water. In a pressure cooker, water is heated at a pressure more than 1atm and temperature above 100°C. Steam below temperature 100°C is called super-cooled vapour.

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Question 593 Marks

Let the temperatures T₁ and T₂ of the two heat reservoirs in an ideal Carnot engine be 1500°C and 500°C respectively. Which of these, increasing T₁ by 100°C or decreasing T₂ by 100°C, would result in a greater improvement in the efficiency of the engine?

Answer

The efficiency of Carnot engine is given by,

$\eta=1-\frac{\text{T}_2}{\text{T}_1}=\frac{\text{T}_1-\text{T}_2}{\text{T}_1}$

When T, is increased from 1500°C to 1600°C or 1600 + 273 = 1873K and T, remains constant i.e., 500°C, or 500 + 273 = 773K, then

$\eta_1=\frac{1873-773}{1873}=\frac{1100}{1873}$

$=\frac{1100\times100}{1873}=58.73\%$

When T₁ remains constant i.e., 1500°C or 1500 + 273 = 1773K and T₂ is decreased by 100 °C i.e., from 500°C to 400°C or 400 + 273 = 673K then

$\eta_2=\frac{1773-673}{1773}=\frac{1100}{1773}$

$=\frac{1100\times100}{1773}=62.04\%$

Thus, $\eta_2>\eta_1$

Hence efficiency will be increased if T₂ is decreased from 500°C to 400°C.

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Question 603 Marks

No real engine can have efficiency greater than that of a Carnot engine working between the same two temperatures, why?

Answer

A Carnot engine is an ideal heat engine from the following points of view:

There is absolutely no friction between the walls of cylinder and the piston.
The working substance is an ideal gas. In a real engine, these conditions cannot be fulfilled and hence no heat engine working between the same two temperatures can have efficiency greater than that of Carnot engine.

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Question 613 Marks

Calculate the efficiency of a Carnot's engine working between steam point and ice point.

Answer

Here, steam point,
$\text{T}_1=100\ ^\circ\text{C}$
$=100+273=373\text{K}$
Ice point , $\text{T}_2=0^\circ{\text{C}}$
$=0+273=273\text{K}$
As $\eta=1-\frac{\text{T}_2}{\text{T}_1}$
$\therefore\ \eta=1-\frac{273}{373}=\frac{100}{373}$
$=\frac{100}{373}\times100\%$
$=26.81\%$

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Question 623 Marks

The temperature of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C. When B and C are mixed, the temperature is 23°C. What would be the temperature when A and C are mixed?

Answer

Let S_A, S_B, S_C be the specific heats of liquids A, B and C respectively.
When A and B are mixed,
$\text{mS}_{\text{A}}(16-12)=\text{mS}_\text{B}(19-16)$
$\text{S}_\text{A}=\frac34\text{S}_\text{B}\dots\text{(i)}$
When B and C are mixed,
$\text{ms}_\text{B}(23-19)=\text{mS}_\text{C}(28-23)$
$\text{S}_\text{B}=\frac{5}{4}\text{S}_\text{C}\dots\text{(ii)}$
From (i) and (ii),
$\text{S}_\text{A}=\frac{15}{16}\text{S}_\text{C}$
Substituting in (iii), we get
$\frac{15}{16}(\theta-12)\text{S}_\text{C}=(28-\theta)\text{S}_\text{C}$ or $\theta=22.0^\circ\text{C}.$

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Question 633 Marks

What is a heat engine? What is the best way to increase efficiency of a heat engine? Is it possible to design a thermal engine that has 100% efficiency?

Answer

A heat engine is a device (or a combination) which converts heat into work.
Its efficiency, $\eta=\frac{\text{Work output}}{\text{Heat input}}$
$\eta=1-\frac{\text{T}_2}{\text{T}_1}$
Where, T₂ = temperature of sink
T₁ = temperature of source.
From above expression, we can see that for 100% efficiency, T₂ = 0
It is impossible to design a thermal engine that has 100% efficiency because it is not possible to have a sink with kelvin temperature.

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Question 643 Marks

A Carnot engine whose heat sink is at 27°C has an efficiency of 40%. By how many degrees should the temperature of source be changed to increase the efficiency by 10% of the original efficiency?

Answer

Here, T₂ = 27°C = 27 + 273 = 300K
$\eta=40\%,\text{ T}_2=?$
From, $\eta=1-\frac{\text{T}_2}{\text{T}_1}$
$\frac{\text{T}_2}{\text{T}_1}=1-\eta$
$=1-\frac{40}{100}$
$=\frac{60}{100}$
$=\frac35$
$\text{T}_1=\frac53\text{t}_2$
$=\frac{5}{3}\times300=500\text{K}$
Increase in efficiency = 10% of 40 = 4%
Let $\text{T}_1'$ be the new temprature of the source.
As $\eta'=1-\frac{\text{T}_2}{\text{T}'_1},$
$\therefore\frac{\text{T}_2}{\text{T}'_1}=1-\eta'$
$=1-\frac{44}{100}$
$=\frac{56}{100}$
$\text{T}'_1=\frac{100}{56}$
$\text{T}_2=\frac{100}{56}\times300$
$=535.7\text{K}$
$\therefore$ Increase in temperature of source = 535.7 - 500 = 35.7K

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Question 653 Marks

A gram molecule of a gas at 127°C expands isothermally until its volume is doubled. Find the amount of work done and heat absorbed.

Answer

Here, temperature of the gas,
T=273 + 127 = 400 K
Let initial volume of the gas,
V₁ = V
$\therefore$ Final volume of the gas,
V₂ = 2V
In an isothermal expansion,
Work done (W) $=2.3026\text{ RT }\log_{10}\frac{\text{V}_2}{\text{V}_1}$
$=2.3026\times8.3\times400\times\log_{10}\frac{2\text{V}}{\text{V}}$
$=2.3026\times8.3\times400\times0.3010$
$\text{W}=2.30\times10^{3}\text{joule}$
If H is the amount of heat absorbed, then
$\text{H}=\frac{\text{W}}{\text{J}}=\frac{2.30\times10^3}{4.2}=548\text{b cal}$

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Question 663 Marks

The efficiency of heat engine cannot be 100%. Explain?

Answer

The efficiency of heat engine is given by,
$\eta=1-\frac{\text{Q}_2}{\text{Q}_1}$
$\text{Q}_2=$ Heat rejected to the sink,
$\text{Q}_1=$ Heat absorbed from the source
$\eta=1$ or 100% if and only if Q₂ = 0
This cannot happen because if Q₂ = 0, then the temperature of the working substance will go on increasing. A stage will come when the temperature of the working substance becomes equal to the temperature of the source. In this situation, there is no transfer of heat from source to the working substance.
Hence, we will not get the output.

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Question 673 Marks

A carnot engine with the cold body temperature of 17°C has 50% efficiency. By how much should the temperature of its hot body be changed to increase the efficiency to 60%? When will its efficiency be 100%?

Answer

$\text{T}_2=273+17=290\text{K},\eta=50\%=\frac12$
As, $\eta=1-\frac{\text{T}_2}{\text{T}_1},\frac12=1-\frac{290}{\text{T}_1}$
$\text{T}_1=580\text{K}$
Again $\eta'=\text{60%}=\frac35$
Thus, $\eta'=1-\frac{\text{T}_2}{\text{T}_1},$ or $\frac35=1-\frac{290}{\text{T}_1'}$
$\text{T}_1'=725\text{K}$
Thus, $\text{T}_1'-\text{T}_1=725\text{K}-580\text{K}=145\text{K}$
For $\eta=1,1-\frac{\text{T}_2}{\text{T}_1}=1$ or $\text{T}_2=0\text{K}$

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Question 683 Marks

Write the expressions for C_v and C_p of a gas in terms of gas constant R and constant $\gamma,$ where $\gamma=\frac{\text{C}_\text{P}}{\text{C}_\nu}$

Answer

We know that C_P - C_v = R
and $\frac{\text{C}_\text{P}}{\text{C}_\nu}=\gamma$
From equation (ii) $\text{C}_\text{P}=\gamma\text{C}_\nu$ and sustituting this value in (i),
We have $\gamma\text{C}_\nu-\text{C}_\nu=\text{R}$
$\Rightarrow\text{C}_\nu=\frac{\text{R}}{(\gamma+1)}$
$\therefore\text{C}_\text{P}=\gamma.\text{C}_\nu=\frac{\gamma\text{R}}{(\gamma-1)}$

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Question 693 Marks

A person weighing 60kg takes in 2000 kcal diet in a day. If this energy were to be used in heating the person without any loss, what would be his rise in temperature? Given specific heat of human body is 0.83 cal g^-1 ^oC^-1

Answer

Here, m = 60kg = 60 × 10³g
$\Delta\text{Q}=2000\text{K cal}=2\times10^6\text{ cal}$
$\Delta\text{Q}=?,\text{C}=0.83\text{cal g}^{-1} {^\circ}\text{C}$
From $\Delta\text{Q}=\text{cm}\Delta\text{T}$
$\Delta\text{T}=\frac{\Delta\text{Q}}{\text{cm}}=\frac{2\times10^6}{0.83\times60\times10^3}$
$=\frac{200}{0.83\times6}=40.16^\circ\text{C}$

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Question 703 Marks

An ideal refrigerator runs between -23°C and 27°C,

Find the heat rejected to atmosphere for every joule of work input.
Also, find heat extracted from cold body.
Find coefficient of performance of the refrigerator.

Answer

Let heat rejected Q₁ = x and W = 1J

Now, Q₂ = Q₁ - W = x - 1

Given, T₁ = 273 + 27 = 300K, T₂ = 273 - 23 = 250K

For an ideal process, $\frac{\text{Q}_2}{\text{Q}_1}=\frac{\text{T}_2}{\text{T}_1}$

$\Rightarrow\frac{\text{x}-1}{\text{x}}=\frac{250}{300}$

$\Rightarrow\text{Q}_1=\text{x}=6\text{J}$

Q₂ = 5J
Coefficient of performance,

$\beta=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}=\frac{250}{300-250}=5$

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Question 713 Marks

Find the pressure required to compress a gas adiabatically at atmospheric pressure to one fifth of its volume $(\gamma=1.4)$

Answer

Here, P₁ = 1atm
Let V₁ = x c.c.;
$\text{V}_2=\frac{\text{x}}{5}\text{ c.c.;}$
$\gamma=1.4;\text{ P}=?$
Using the relation, $\text{P}_1\text{V}^\gamma_1=\text{P}_2\text{V}^\gamma_2$
$\therefore\text{P}_2=\text{P}_1\Big(\frac{\text{V}_1}{\text{V}_2}\Big)^\gamma=1\bigg(\frac{\text{x}}{\frac{\text{x}}{5}}\bigg)^{1.4}=(5)^{1.4}$
Taking $\log$ both sides, we get
$\log\text{P}_2=1.4\log5$
$=1.4\times0.6990=0.97860$
$\therefore\text{P}_2=9.519\text{ atm}.$

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Question 723 Marks

Differentiate between evaporation and boiling.

Answer

Evaporation is a slow process from the liquid to the gaseous state which takes place at the surface of a liquid and at all temperatures. Boiling is a rapid change of a substance from the liquid to the gaseous state which takes place throughout the mass of the liquid at a definite temperature.

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Question 733 Marks

State Dulong and Petit's law.

Answer

According to the Dulong and Petit's law, the specific heat per mole of a chemically pure crystalline solid is approximately 6 cal mol^-1K^-1 or roughly 25J mole^-1 K^-1. This law is obeyed with quite a good approximation for many substances at the room temperature.

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Question 743 Marks

Establish relation between two specific heats of a gas. Which is greater and why?

Answer

Relation between C_pand C_v.
Suppose one mole of a gas is heated so that its temperature rises by dT. Heat supplied = 1 × C_p × dT = C_v dT ...(i)
Since the volume is constant, the gas will not perform external work in accordance with the first law of thermodynamics and the heat supplied will be just equal to the increase in the internal energy of the gas.
dU = C_vdT ...(ii)
Let the gas be heated at constant pressure to again increase its temperature by dT, and dQ be the amount of heat supplied, therefore,
$\therefore$ dQ = 1 × C_p × dT = C_p dT ...(iii)
The heat supplied at a constant pressure increases the temperature by dT hence, increases its internal energy by dU as well as enables the gas to perform work dW.
dW = PV ...(iv)
From the first law of thermodynamics, we have,
dQ = dU + dW
Substituting the values, we get,
C_pdT = C_v dT + PDV
But PV = RT (For one mole of the gas)
Or PdV = RdT
$\therefore$ C_pdT = C_v dT + RdT
or C_p - C_v = R ...(v)
This is the relation between two principal specific heats of the gas when C_p, C_v and R are measured in the units of either heat or of work.
C_p > C_v because a part of the energy supplied in the adiabatic process goes to increase the volume of the gas and the remaining increases the temperature.

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Question 753 Marks

Calculate the efficiency of Carnot's engine working between steam point and ice point.

Answer

Here, steam point $\text{T}_1=100^\circ\text{C}$
$=100+273=373\text{K}$
and ice point $\text{T}_2=0^\circ\text{C}$
$=0+273=273\text{K}$
$\therefore\eta=1-\frac{\text{T}_2}{\text{T}_1}$
$=1-\frac{273}{373}=\frac{100}{373}$
$\therefore\eta=\frac{100}{373}\times100\%=26.81\%$

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Question 763 Marks

A refrigerator transfers 250J heat per second from -23°C to 25°C. Find the power consumed, assuming no loss of energy.

Answer

Here, Q₂ = 250J s^-1
T₂ = -23°C = -23 + 273 = 250K
T₁ = 25°C = 25 + 273 = 298K
We know, $\beta=\frac{\text{Q}_2}{\text{W}}=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}$
$\therefore\text{W}=\frac{\text{Q}_2(\text{T}_1-\text{T}_2)}{\text{T}_2}$
$\text{W}=\frac{250(298-250)}{250}=\frac{250\times48}{250}$
$\text{W}=48\text{J s}^{-1}$

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Question 773 Marks

A Carnot engine absorbs 1000J of heat energy from a reservoir at 12.7°C and rejects 600J of heat energy during each cycle. Calculate. (i) efficiency of the engine, (ii) temperature of sink, (iii) amount of useful work done per cycle.

Answer

Here, $\text{Q}-1=1000\text{J},\text{Q}_2=600\text{J},$
$\text{T}_1=127^\circ\text{C}=127+273=400\text{K}$
$\eta=?,\ \text{T}_2=?,\ \text{W}=?$
From $\frac{\text{Q}_2}{\text{Q}_1}=\frac{\text{T}_2}{\text{T}_2},\text{T}_2=\frac{\text{Q}_2}{\text{Q}_1}\times\text{T}_1$
$\text{T}_2=\frac{600}{1000}\times400=240\text{K}$
$=240-273=-33^\circ\text{C}$
$\eta=1-\frac{\text{T}_2}{\text{T}_1}=1-\frac{240}{400}=0.4=40\%$
Also, $\text{W}=\text{Q}_1-\text{Q}_2=1000-600=400\text{J}$

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Question 783 Marks

Absolute zero temperature is not the temperature of zero energy. Explain.

Answer

At absolute zero, the energy of translatory motion of molecules ceases but the other forms of energy such as inter-molecular, potential energy of molecular motion, etc. do not become zero. Therefore, absolute temperature is not the temperature of zero energy.

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Question 793 Marks

Consider one mole of perfect gas in a cylinder of unit cross-section with a piston attached (figure). A spring (spring constant k) is attached (unstretched length L) to the piston and to the bottom of the cylinder. Initially, the spring is unstretched and the gas is in equilibrium.
A certain amount of heat Q is supplied to the gas causing an increase of volume from V₀ to V₁.

What is the initial pressure of the system?
What is the final pressure of the system?
Using the first law of thermodynamics, write down a relation between Q, P_a, V, V₀, and k.

Answer

p_i = p_a
$\text{p}_\text{f}=\text{p}_\text{a}+\frac{\text{k}}{\text{A}}(\text{V}-\text{V}_0)$
According to first law of thermodynamics,

$\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$ Where,

$\Delta\text{U}=\text{C}_\text{V}(\text{T}-\text{T}_0),$ $\Delta\text{W}=\text{p}_\text{a}(\text{V}-\text{V}_0)+\frac12\text{k}(\text{V}-\text{V}_0)^2$

$\Delta\text{Q}=\text{p}_\text{a}(\text{V}-\text{V}_0)+\frac12\text{k}(\text{V}-\text{V}_0)^2+\text{C}_\text{V}(\text{T}-\text{T}_0)$

Where, $\text{T}_0=\text{p}_\text{a}\frac{\text{V}_0}{\text{R}},$

$\Rightarrow\text{T}=\Big[\text{p}_\text{a}+\Big(\frac{\text{K}}{\text{A}}\Big)\times(\text{V}-\text{V}_0)\Big]\frac{\text{V}}{\text{R}}$

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Question 803 Marks

The initial state of a certain gas is (P_i , V_i , T_i ). It undergoes expansion till its volume becoms V_f . Consider the following two cases:

The expansion takes place at constant temperature.
The expansion takes place at constant pressure.

Plot the P-V diagram for each case. In which of the two cases, is the work done by the gas more?

Answer

The expension from V_i to Vf tempreature T_i remains constant so isothermal expension i.e. P_iV_i = P_fV_f constant T.
The expension is at constant pressure p_i so isobaric process so graph P-V will be parallel to V axis till its volume becomes V_f As the area enclosed by graph (a) is less than (b) with volume axis so W.D. by process (b) is more than of (a).

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Question 813 Marks

Find out whether these phenomena are reversible or not:

Waterfall.
Rusting of iron.

Answer

Waterfall: The falling of water cannot be reversible process. During the water fall, its potential energy convert into kinetic energy of the water.

On striking the ground, some part of potential energy converts into heat and (sound not possible that heat and the sound). In nature, it automatically convert the kinetic energy and potential energy so that the water will rise back so waterfall is not a reversible process.

Rusting of iron: In rusting of iron, the iron become oxidised with the oxygen of the air as it is a chemical reaction, it cannot be reversed.

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Question 823 Marks

Prove that the slope of P-V graph for an adiabatic process is $\gamma$ times that of the isothermal process.

Answer

For isothermal process, PV = constant

Differentiating, VdP + PDV = 0

$\therefore\frac{\text{dP}}{\text{dV}}=-\frac{\text{P}}{\text{V}}$

For adiabatic process, $\text{PV}^{\gamma}=$ constant.

Differentiating,

$\text{V}^{\gamma}\text{dP}+\gamma\text{PV}^{\gamma-1}\text{dV}=0$

$\therefore\frac{\text{dP}}{\text{dV}}=-\frac{\gamma\text{P}}{\text{V}}$

Comparing the two ratios, we can say, slope of adiabatic process is $\gamma$ times the slope of isothermal process.

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Question 833 Marks

The efficiency of a Carnot engine is $\frac12.$ If the sink temperature is reduced by 100°C, then engine efficiency becomes $\frac23.$ Find

Sink temperature.
Source temperature.
Explain, why a Carnot engine cannot have 100% efficiency?

Answer

Efficiency, $\eta=1-\frac{\text{T}_2}{\text{T}_1}$

Where, T₂ = sink temperature

T₁ = source temperature

$1-\frac{\text{T}_2-100}{\text{T}_1}\dots\text{(i)}$

$1-\Big(\frac{\text{T}_2-100}{\text{T}_1}\Big)=\frac23\dots\text{(ii)}$

From equation (i), $\frac{\text{T}_2}{\text{T}_1}=\frac{1}{2}$ and equation (ii),

${\text{T}_2-100\over\text{T}_1}=\frac32$

$\Rightarrow\text{T}_2=300\text{K}$

Substituting in equation (i), T₁ = 600K
As efficiency, $\eta_2\Rightarrow1-\frac{\text{T}_2}{\text{T}_1}$

$\therefore$ It equals to 1 only when $\frac{\text{T}_2}{\text{T}_1}=0$ or $\text{T}_2=0\text{K}$

But absolute zero is not possible.

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3 Marks Question - Page 2 - Physics STD 11 Science Questions - Vidyadip