A monoatomic ideal gas at pressure $P$ and volume $V$ is first adiabatically compressed to volume $V...

Adiabatic processFor an adiabatic process involving an ideal gas, we use the relation:$$ P V^\gamma=\text { constant } $$Where symbols hold their normal meanings. For a monoatomic ideal gas, $\gamma=\frac{5}{3}$.Given the initial state $(P, V)$ and the final volume after adiabatic compression $V_f=\frac{V}{8}$, the final pressure $P_f$ can be found using:$$ P V^\gamma=P_f\left(\frac{V}{8}\right)^\gamma $$Solving for $P_f$ :$$ P_f=P\left(\frac{V}{\frac{V}{8}}\right)^\gamma=P(8)^\gamma=P \cdot 8^{5 / 3} $$Isothermal processIn an isothermal process, the temperature remains constant, so the ideal gas law $P V=n R T$ applies and the product of pressure and volume remains constant.Let the pressure after isothermal expansion back to the original volume $V$ be $P_f{ }^{\prime}$. The initial state before isothermal expansion is $\left(P_f, \frac{V}{8}\right)$ and the final state is $\left(P_f^{\prime}, V\right)$.From the ideal gas law:$$ P_f \cdot \frac{V}{8}=P_f^{\prime} \cdot V $$Solving for $P_f{ }^{\prime}$ :$$ P_f^{\prime}=\frac{P_f \cdot \frac{V}{8}}{V}=\frac{P_f}{8} $$Substituting $P_f=P \cdot 8^{5 / 3}$ into the equation:$$ P_f^{\prime}=\frac{P \cdot 8^{5 / 3}}{8}=P \cdot 8^{5 / 3-1}=P \cdot 8^{2 / 3} $$So, the final pressure $P_f{ }^{\prime}$ is: $P_f{ }^{\prime}=P \cdot 8^{2 / 3}$Thus, the final pressure of the gas after the entire process is: $P_f{ }^{\prime}=P \cdot 8^{2 / 3}$$$ P_f^{\prime}=4 P $$

One mole of an ideal gas of volume $V$ and temperature $T$ is allowed to expand adiabatically to vol...

Given the following initial conditions for an ideal gas:Initial volume $ V $Final volume $ 2V $Initial temperature $ T $Number of moles $ n = 1 $The process described is adiabatic, meaning no heat is exchanged with the surroundings. Furthermore, it's specified that no external work is done, indicating a free expansion, where the gas expands into a vacuum and performs no work on its surroundings. As a result, because it is adiabatic and no work is done, the internal energy and temperature of the gas remain constant throughout the process. Consequently, the temperature after expansion remains $ T $.Using the ideal gas law, $ PV = nRT $, we analyze the initial and final states:Initial State:$ P_iV = RT $Final State:In the final state, the volume is $ 2V $ and the temperature is still $ T $:$ P_f \cdot 2V = RT $To find the final pressure $ P_f $, rearrange the equation:$ P_f = \frac{RT}{2V} $Thus, the pressure of the gas after expansion is $\frac{RT}{2V}$.

Two identical boxes contain the same ideal gas. Let $\left(n_1, \lambda_1, T_1\right)$ and $\left(n_...

To find the mean free path and the temperature when one gas box is emptied into the other, we can start with the known formula for mean free path:$ \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} $Where:$ k_B $ is the Boltzmann constant$ d $ is the diameter of the gas molecules$ T $ is the absolute temperature of the gas$ P $ is the pressure of the gasUsing the ideal gas law:$ PV = nRT \quad \Rightarrow \quad P = \frac{nRT}{V} $We can rewrite:$ \lambda = \frac{k_B V}{\sqrt{2} \pi d^2 nR} $Given that both boxes are identical, each with volume $ V_1 = V_2 $ and that the gas molecules have the same diameter $ d_1 = d_2 $, we introduce a constant for convenience:$ \frac{k_B V}{\sqrt{2} \pi d^2 R} = K $Thus, for each box:$ \lambda = \frac{K}{n} $$ \lambda_1 = \frac{K}{n_1}, \quad \lambda_2 = \frac{K}{n_2} $This gives:$ n_1 = \frac{K}{\lambda_1}, \quad n_2 = \frac{K}{\lambda_2} $Therefore, for the combined system:$ n_{\text{net}} = n_1 + n_2 = \frac{K}{\lambda_{\text{net}}} $This leads to:$ \frac{1}{\lambda_1} + \frac{1}{\lambda_2} = \frac{1}{\lambda_{\text{net}}} $Regarding the temperature, the total internal energy is conserved. For an ideal gas, the net internal energy is expressed as:$ U = \frac{3}{2} nRT $For the two boxes:$ U_1 = \frac{3}{2} n_1 R T_1 \quad \text{and} \quad U_2 = \frac{3}{2} n_2 R T_2 $So, the final internal energy $ U_f $ after mixing is:$ U_1 + U_2 = U_f $Cancelling out $\frac{3}{2} R$:$ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_f $Thus, the final temperature $ T_f $ is given by:$ T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} $

A metallic cube initially kept at a temperature $T$ is emitting black body radiation with a power $P...

Given:A metallic cube (edge = $ L $) radiates as a black body.Initial temperature = $ T $Power emitted = $ P $When $ T $ increases by $ 1\% $, $ P $ increases by $ 4.5\% $We are asked:→ What is thepercentage increase in volumeof the cube in this process?Step 1: Expression for power radiatedFor a black body (Stefan–Boltzmann law):$ P = e\sigma A T^4 $where$ e $ = emissivity,$ \sigma $ = Stefan–Boltzmann constant,$ A $ = surface area,$ T $ = temperature.For a cube of edge $ L $:$ A = 6L^2 $Thus,$ P = 6 e \sigma L^2 T^4 $Step 2: Take differentials / fractional changesTake logarithm and differentiate:$ \frac{dP}{P} = 2 \frac{dL}{L} + 4 \frac{dT}{T} $This gives:$ \frac{\Delta P}{P} \approx 2 \frac{\Delta L}{L} + 4 \frac{\Delta T}{T} $Step 3: Substitute the given fractional changesGiven:$ \frac{\Delta T}{T} = 1\% = 0.01 $and$ \frac{\Delta P}{P} = 4.5\% = 0.045 $Plug in:$ 0.045 = 2 \frac{\Delta L}{L} + 4(0.01) $$ 0.045 = 2 \frac{\Delta L}{L} + 0.04 $$ 2 \frac{\Delta L}{L} = 0.005 $$ \frac{\Delta L}{L} = 0.0025 = 0.25\% $Step 4: Percentage change in volumeFor a cube:$ V = L^3 \Rightarrow \frac{\Delta V}{V} = 3 \frac{\Delta L}{L} $Hence:$ \frac{\Delta V}{V} = 3(0.0025) = 0.0075 = 0.75\% $✅Final Answer:Percentage increase in volume = 0.75%✅ Correct Option:(A) 0.75%

Heat And Thermodynamics - IAT-IISER Physics PYQs

Q: Two identical containers kept at the same temperature carry the same gas with different molecular me...

The mean free path ($\lambda$) of a gas molecule is given by the formula:$ \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} $where:$ k_B $ is the Boltzmann constant,$ T $ is the absolute temperature of the gas,$ d $ is the diameter of the gas molecules,$ P $ is the pressure of the gas.In this scenario:Both containers are at the same temperature: $ T_1 = T_2 = T $.The same gas implies: $ d_1 = d_2 = d $.For simplification, define:$ k = \frac{k_B T}{\sqrt{2} \pi d^2} $This leads to:$ l_1 = \frac{k}{P_1}, \quad l_2 = \frac{k}{P_2} \Rightarrow P_1 = \frac{k}{l_1}, \quad P_2 = \frac{k}{l_2} $The mean free path of the resultant mixture, $ l_{\text{net}} $, can be expressed in terms of the net pressure ($ P_{\text{net}} = P_1 + P_2 $):$ l_{\text{net}} = \frac{k}{P_{\text{net}}} $Thus, substituting for $ P_{\text{net}} $:$ l_{\text{net}} = \frac{k}{\frac{k}{l_1} + \frac{k}{l_2}} = \frac{1}{\frac{1}{l_1} + \frac{1}{l_2}} $Finally, simplify this expression to find:$ l_{\text{net}} = \frac{l_1 l_2}{l_1 + l_2} $

Q: An ideal rigid diatomic gas undergoes expansion at constant pressure when $\Delta Q$ heat is added t...

Let's work through the problem step by step.For an ideal gas, the first law of thermodynamics states:$$\Delta Q = \Delta U + W$$In an isobaric (constant pressure) process, some of the heat added goes into doing work by expanding the gas, while the rest increases the internal energy.For an ideal gas, the change in internal energy is given by:$$\Delta U = n C_v \Delta T$$And the heat added is:$$\Delta Q = n C_p \Delta T$$where $ C_v $ and $ C_p $ are the molar heat capacities at constant volume and constant pressure, respectively.Thus, the ratio is:$$\frac{\Delta U}{\Delta Q} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p}$$For a rigid diatomic gas (one where the vibrational modes are not active), the degrees of freedom are:Translational: 3Rotational: 2This means the molar internal energy is:$$U = \frac{5}{2}RT$$so:$$C_v = \frac{5}{2}R$$Since $ C_p = C_v + R $, it follows that:$$C_p = \frac{5}{2}R + R = \frac{7}{2}R$$Now, the ratio becomes:$$\frac{C_v}{C_p} = \frac{\frac{5}{2}R}{\frac{7}{2}R} = \frac{5}{7}$$Thus, the ratio of the change in the internal energy to the heat supplied is:$$\frac{\Delta U}{\Delta Q} = \frac{5}{7}.$$The correct answer is Option B.

Q: Consider a Carnot heat engine operating between a heat reservoir at temperature 600 K , and an exter...

The efficiency $\eta$ of a Carnot heat engine is given by:$ \eta = 1 - \frac{T_c}{T_h} $where $T_h$ is the temperature of the heat reservoir and $T_c$ is the temperature of the atmosphere.Substituting the given temperatures:$ \eta = 1 - \frac{300}{600} = 0.5 $The work done by the heat engine ($W_{\text{HE}}$) is calculated as:$ W_{\text{HE}} = \eta Q_h $Given that $Q_h = 1000 \, \text{kJ}$:$ W_{\text{HE}} = 0.5 \times 1000 = 500 \, \text{kJ} $The work done by the heat engine $W_{\text{HE}}$ is utilized as input work $W_{\text{ref}}$ for the refrigerator.The Coefficient of Performance (COP) of the refrigerator is defined as:$ \text{COP} = \frac{T_L}{T_H - T_L} $where $T_L$ is the temperature at which the refrigerator operates, and $T_H$ is the temperature of the atmosphere.Inserting the given values:$ \text{COP} = \frac{200}{300 - 200} = 2 $The heat extracted from the cold reservoir $Q_L$ by the refrigerator is:$ Q_L = \text{COP} \times W_{\text{ref}} $Substituting $W_{\text{ref}} = 500 \, \text{kJ}$:$ Q_L = 2 \times 500 = 1000 \, \text{kJ} $Total heat released into the atmosphere is calculated as follows:The heat released by the heat engine ($Q_c$) is:$ Q_c = Q_h - W_{\text{HE}} $Substitute the values:$ Q_c = 1000 - 500 = 500 \, \text{kJ} $The heat released by the refrigerator ($Q_H$) is:$ Q_H = Q_L + W_{\text{ref}} $Substitute the values:$ Q_H = 1000 + 500 = 1500 \, \text{kJ} $Thus, the total heat released into the atmosphere by the combined system is:$ Q_{\text{total}} = Q_c + Q_H $$ Q_{\text{total}} = 500 + 1500 = 2000 \, \text{kJ} $

01

PYQ 2020

easy

physics ID: iat-iise

Two identical containers kept at the same temperature carry the same gas with different molecular mean free paths and . The gas from the first container is completely transferred into the second at the same temperature. What is the molecular mean free path of the resultant mixture in the second container?

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2

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4

02

PYQ 2020

easy

physics ID: iat-iise

An ideal rigid diatomic gas undergoes expansion at constant pressure when heat is added to it. What is the ratio of the change in the internal energy to the heat supplied ?

1

2

3

4

03

PYQ 2022

easy

physics ID: iat-iise

A monoatomic ideal gas at pressure and volume is first adiabatically compressed to volume and then is allowed to expand isothermally back to its original volume. What is the final pressure of the gas?

1

2

3

4

04

PYQ 2023

easy

physics ID: iat-iise

Consider a Carnot heat engine operating between a heat reservoir at temperature 600 K , and an external atmosphere at temperature 300 K . In one cycle, 1000 kJ of heat is extracted from the heat reservoir, and the associated work is input to a reversible refrigerator that operates between 200 K and the same external atmosphere at 300 K . The refrigerator completes one cycle and releases heat into the atmosphere. How much heat is released into the atmosphere at the end of one cycle of the combined system?

1

2500 Kj

2

2000 kJ

3

1000 kJ

4

500 kJ

05

PYQ 2024

easy

physics ID: iat-iise

One mole of an ideal gas of volume and temperature is allowed to expand adiabatically to volume while doing no external work. The universal gas constant is . What is the pressure of the gas after expansion?

1

2

3

4

06

PYQ 2024

easy

physics ID: iat-iise

Two identical boxes contain the same ideal gas. Let and be the number density, mean free path and temperature of the gas in the first and the second box, respectively. One of the boxes is empty into the other one. What will be the mean free path? and temperature of the gas now?

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07

PYQ 2025

easy

physics ID: iat-iise

A metallic cube initially kept at a temperature is emitting black body radiation with a power (energy emitted per unit time). If is increased by, the power being radiated increases by . What is the approximate percentage increase in the volume of the cube in this process?

1

2

3

$ $

4

$ $

Heat And Thermodynamics

High-Yield Trend

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Heat And Thermodynamics

High-Yield Trend

Questions 7 MCQs

Official Solution

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Questions
7 MCQs