LEVELBoard

Animated Solution for Physics - Thermodynamics: In a given process of an ideal gas, $d W = 0$ and $d Q < 0$ . Then for the gas

Q: Why does $dW = 0$ mean the volume is constant?

Work done by a gas is given by $dW = p dV$. For the work to be zero while pressure exists, the change in volume $dV$ must be zero. This means the container is rigid.

Q: Can internal energy change if no work is done?

Absolutely! Internal energy can change through heat transfer. If you heat a rigid container, no work is done, but the gas gets hotter and its internal energy increases.

Q: Why does internal energy depend only on temperature for an ideal gas?

In an ideal gas, we assume there are no intermolecular forces. Therefore, there is no potential energy between molecules. All internal energy is purely the kinetic energy of the molecules, which is directly proportional to temperature.

Q: What if the gas was not ideal?

For a real gas, internal energy depends on both temperature and volume because intermolecular forces contribute to potential energy. However, for JEE problems, unless specified, we assume ideal behavior.

Q: Does $dQ < 0$ always mean temperature decreases?

Not necessarily! If the gas is being compressed ($dW < 0$) and the work done on the gas is greater than the heat lost, the internal energy (and temperature) could actually increase. You must always evaluate the full First Law equation.

Select Answer:

Visualized Solution

$d W = 0$

$d W = 0$
$⟹ d V = 0$

$d Q < 0$

$d Q < 0$
$⟹ Heat is rejected$

$First Law of Thermodynamics$

$d Q = d U + d W$

$Substituting values$

$d Q = d U + 0$
$⟹ d Q = d U$

$Sign of d U$

$∵ d Q < 0$
$∴ d U < 0$

$Conclusion$

$U = \frac{f}{2} n R T$
$⟹ U \propto T$
$∵ d U < 0 ⟹ d T < 0$

$What if?$

$If d W > 0 and d Q = 0$
$⟹ d U = - d W < 0$

00:00 / 00:00

The Sigma Insight: First Law of Thermodynamics

Solution Diagram

Analyzing the Setup

Imagine a gas trapped inside a cylinder with a piston. The problem explicitly states that the work done by the gas,

d W

, is exactly zero.

This is a crucial constraint! Physically, this means the piston is completely fixed in place, and the volume of the gas cannot change.

We are also given that the heat exchange,

d Q

, is less than zero. This indicates that heat is not being added to the system.

Instead, heat is being extracted or lost to the surroundings, meaning the gas is cooling down.

The Master Equation

To connect heat, work, and internal energy, we must bring in the First Law of Thermodynamics.

This fundamental law states that the heat supplied to a system equals the change in its internal energy plus the work done by the system.

Mathematically, we write this as:

d Q = d U + d W

Let's substitute our known constraints into this master equation. Since the work done

d W

is zero, the equation simplifies beautifully.

d Q = d U + 0

This leaves us with a direct relationship: the heat exchange

d Q

is exactly equal to the change in internal energy

d U

Final Calculation

Now, remember our second constraint:

d Q

is negative.

Since

d Q

equals

d U

, the change in internal energy

d U

must also be negative.

d U < 0

This means the final internal energy is strictly less than the initial internal energy.

For an ideal gas, the internal energy depends strictly on its absolute temperature. The formula is

U = \frac{f}{2} n R T

Therefore, a decrease in internal energy directly implies a decrease in temperature.

U \propto T ⟹ d T < 0

So, we can confidently conclude that the temperature will decrease.

Common Traps & FAQs

Why does

d W = 0

mean the volume is constant?

Work done by a gas is given by

d W = p d V

. For the work to be zero while pressure exists, the change in volume

d V

must be zero. This means the container is rigid.

Can internal energy change if no work is done?

Absolutely! Internal energy can change through heat transfer. If you heat a rigid container, no work is done, but the gas gets hotter and its internal energy increases.

Why does internal energy depend only on temperature for an ideal gas?

In an ideal gas, we assume there are no intermolecular forces. Therefore, there is no potential energy between molecules. All internal energy is purely the kinetic energy of the molecules, which is directly proportional to temperature.

What if the gas was not ideal?

For a real gas, internal energy depends on both temperature and volume because intermolecular forces contribute to potential energy. However, for JEE problems, unless specified, we assume ideal behavior.

Does

d Q < 0

always mean temperature decreases?

Not necessarily! If the gas is being compressed (

d W < 0

) and the work done on the gas is greater than the heat lost, the internal energy (and temperature) could actually increase. You must always evaluate the full First Law equation.

Similar Questions

LEVELJEE Main

For an ideal gas

* Multiple Correct Options

(A)

the change in internal energy in a constant pressure process from temperature

T_{1}

T_{2}

is equal to

n C_{V} (T_{2} - T_{1})

, where

C_{V}

is the molar heat capacity at constant volume and

n

the number of moles of the gas

(B)

the change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process

(C)

the internal energy does not change in an isothermal process

(D)

no heat is added or removed in an adiabatic process

JEE Advanced 2014

LEVELJEE Main

An ideal gas in thermally insulated vessel at internal pressure = $P_{1}$ , volume = $V_{1}$ and absolute temperature = $T_{1}$ expands irrversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are $P_{2}$ , $V_{2}$ and $T_{2}$ , respectively. For this expansion,

* Multiple Correct Options

(A)

q = 0

(B)

T_{2} = T_{1}

(C)

P_{2} V_{2} = P_{1} V_{1}

(D)

P_{2} V_{2}^{γ} = P_{1} V_{1}^{γ}

JEE Advanced 2017

LEVELJEE Advanced

An ideal gas is expanded from $(p_{1}, V_{1}, T_{1})$ to $(p_{2}, V_{2}, T_{2})$ under different conditions. The correct statement(s) among the following is(are)

* Multiple Correct Options

(A)

The work done on the gas is maximum when it is compressed irreversibly from

(p_{2}, V_{2})

(p_{1}, V_{1})

against constant pressure

p_{1}

(B)

The work done on the gas is less when it is expanded reversibly from

V_{1}

V_{2}

under adiabatic conditions as compared to that when expanded reversibly from

V_{1}

V_{2}

under isothermal conditions.

(C)

The change in internal energy of the gas (i) zero, if it is expanded reversibly with

T_{1} = T_{2}

, and (ii) positive, if it is expanded reversibly under adiabatic conditions with

T_{1} \neq = T_{2}

(D)

If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic.

LEVELJEE Main

An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If $T_{i}$ is the initial temperature and $T_{f}$ is the final temperature, then which of the following statements is correct ?

(A)

(T_{f})_{i r r e v} > (T_{f})_{r e v}

(B)

T_{f} > T_{i}

for reversible process but

T_{f} = T_{i}

for irreversible process

(C)

(T_{f})_{r e v} = (T_{f})_{i r r e v}

(D)

T_{f} = T_{i}

for both reversible and irreversible processes

JEE Main 2019

LEVELJEE Main

Following figure shows two processes A and B for a gas. If $Δ Q_{A}$ and $Δ Q_{B}$ are the amount of heat absorbed by the system in two cases, and $Δ U_{A}$ and $Δ U_{B}$ are changes in internal energies respectively, then

(A)

Δ Q_{A} > Δ Q_{B}, Δ U_{A} > Δ U_{B}

(B)

Δ Q_{A} < Δ Q_{B}, Δ U_{A} < Δ U_{B}

(C)

Δ Q_{A} > Δ Q_{B}, Δ U_{A} = Δ U_{B}

(D)

Δ Q_{A} = Δ Q_{B}, Δ U_{A} = Δ U_{B}

JEE Main 2020

LEVELJEE Main

For one mole of an ideal gas, which of these statements must be true? (A) $U$ and $H$ each depends only on temperature. (B) Compressibility factor $Z$ is not equal to $1$ . (C) $C_{p, m} - C_{V, m} = R$ (D) $d U = C_{V} d T$ for any process.

An ideal gas undergoes a four step cycle as shown in the P – V diagram below. During this cycle, heat is absorbed by the gas in

A reversible cyclic process for an ideal gas is shown below. Here, P , V and T are pressure , volume and temperature , respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

* Multiple Correct Options

(A)

q_{AC} = Δ U_{BC}

and

w_{AB} = P_{2} (V_{2} - V_{1})

(B)

w_{BC} = P_{2} (V_{2} - V_{1})

and

q_{BC} = Δ H_{AC}

(C)

Δ H_{CA} < Δ U_{CA}

and

q_{AC} = Δ U_{BC}

(D)

q_{BC} = Δ H_{AC}

and

Δ H_{CA} > Δ U_{CA}

JEE Advanced 2019

LEVELJEE Advanced

Comprehension Passage

In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by

T Δ X

, where

T

is temperature of the system and

Δ X

is the infinitesimal change in a thermodynamic quantity

X

of the system. For a mole of monatomic ideal gas

X = \frac{3}{2} R ln (\frac{T}{T _{A}}) + R ln (\frac{V}{V _{A}})

. Here,

R

is gas constant,

V

is volume of gas,

T_{A}

and

V_{A}

are constants. The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities. List-I (I) Work done by the system in process

1 \to 2 \to 3

(II) Change in internal energy in process

1 \to 2 \to 3

(III) Heat absorbed by the system in process

1 \to 2 \to 3

(IV) Heat absorbed by the system in process

1 \to 2

List-II (P)

\frac{1}{3} R T_{0} ln 2

(Q)

\frac{1}{3} R T_{0}

(R)

R T_{0}

(S)

\frac{4}{3} R T_{0}

(T)

\frac{1}{3} R T_{0} (3 + ln 2)

(U)

\frac{5}{6} R T_{0}

Question 1:

If the process carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0}$ , the correct match is,

(A)

\to

Q, II

\to

R, III

\to

P, IV

\to

(B)

\to

S, II

\to

R, III

\to

Q, IV

\to

(C)

\to

Q, II

\to

R, III

\to

S, IV

\to

(D)

\to

Q, II

\to

S, III

\to

R, IV

\to

Question 2:

If the process on one mole of monatomic ideal gas is an shown is as shown in the TV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0}$ , the correct match is

(A)

\to

S, II

\to

T, III

\to

Q, IV

\to

(B)

\to

P, II

\to

R, III

\to

T, IV

\to

(C)

\to

P, II

\to

T, III

\to

Q, IV

\to

(D)

\to

P, II

\to

R, III

\to

T, IV

\to

JEE Advanced 2011

LEVELJEE Advanced

One mole of a monatomic ideal gas is taken through a cycle $A B C D A$ as shown in the $p - V$ diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.

List-I

(P)

Process

A \to B

(Q)

Process

B \to C

(R)

Process

C \to D

(S)

Process

D \to A

List-II

(1)

Internal energy decreases

(2)

Internal energy increases

Work is done on the gas

Animated Solution for Physics - Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a given process of an ideal gas, dW=0 and dQ<0. Then for the gas

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dW=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dQ<0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Law of Thermodynamics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sign of dU

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if?

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

For an ideal gas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An ideal gas is expanded from (p1​,V1​,T1​) to (p2​,V2​,T2​) under different conditions. The correct statement(s) among the following is(are)

An ideal gas undergoes a four step cycle as shown in the P – V diagram below. During this cycle, heat is absorbed by the gas in

A reversible cyclic process for an ideal gas is shown below. Here, P , V and T are pressure , volume and temperature , respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the process carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with P0​V0​=31​RT0​, the correct match is,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the process on one mole of monatomic ideal gas is an shown is as shown in the TV-diagram with P0​V0​=31​RT0​, the correct match is

List-I

List-II

Animated Solution for Physics - Thermodynamics: In a given process of an ideal gas, $d W = 0$ and $d Q < 0$ . Then for the gas

$d W = 0$

$d Q < 0$

$First Law of Thermodynamics$

$Substituting values$

$Sign of d U$

$Conclusion$

$What if?$

An ideal gas is expanded from $(p_{1}, V_{1}, T_{1})$ to $(p_{2}, V_{2}, T_{2})$ under different conditions. The correct statement(s) among the following is(are)

If the process carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0}$ , the correct match is,

If the process on one mole of monatomic ideal gas is an shown is as shown in the TV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0}$ , the correct match is