LEVELJEE Main

Animated Solution for Physics - Thermodynamics: A container with insulating walls is divided into two equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressure $p$ and temperature $T$ , whereas the other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be

Q: Can we use $p_1 V_1^\gamma = p_2 V_2^\gamma$ since it's an adiabatic process?

No! The equation $pV^\gamma = \text{constant}$ is only valid for *reversible* adiabatic processes. Free expansion is a highly irreversible process, so you cannot use this relation. You must rely on the First Law of Thermodynamics directly.

Select Answer:

Visualized Solution

$I ni t ia l S e t u p$

Insulated container divided into two equal parts.
Left part: Ideal gas at $p, V, T$ .
Right part: Vacuum.

$F r ee E x p an s i o n$

Valve is opened.
Gas expands into the vacuum.
This process is called Free Expansion.

$F i r s t L a w o f T h er m o d y nami cs$

Insulating walls $\Rightarrow Q = 0$
Expansion into vacuum $\Rightarrow W = 0$
First Law: $Δ U = Q - W$

$C han g e in I n t er na l E n er g y$

$Δ U = 0 - 0 = 0$
Internal energy $U$ remains constant.

$F ina l T e m p er a t u r e$

For an ideal gas, $U$ depends only on $T$ .
Since $Δ U = 0 \Rightarrow Δ T = 0$
Final temperature, $T_{f} = T$

$B oy l e^{'} s L a w$

Since $T$ is constant, the process is isothermal.
Apply Boyle's Law: $p_{i} V_{i} = p_{f} V_{f}$

$F ina l P r ess u r e$

Initial volume, $V_{i} = V$
Final volume, $V_{f} = V + V = 2 V$
$p \cdot V = p_{f} \cdot (2 V)$

$F ina l A n s w er$

$p_{f} = \frac{p \cdot V}{2 V}$
$p_{f} = \frac{p}{2}$
Final state: $(\frac{p}{2}, T)$

$T h e W a y F or w a r d$

What if the gas was a real gas?
Real gases experience intermolecular forces.
Free expansion of a real gas leads to a slight temperature drop (Joule-Thomson effect).

00:00 / 00:00

The Sigma Insight: First Law of Thermodynamics

Solution Diagram

Analyzing the Setup

Imagine a perfectly insulated container divided right down the middle by a partition with a valve. On the left side, we have an ideal gas happily bouncing around at a pressure

p

, volume

V

, and temperature

T

. On the right side, there is absolutely nothing—a perfect vacuum.

Suddenly, the valve is opened. The gas rushes in to fill the empty space. This specific process, where a gas expands into a vacuum, is known in physics as Free Expansion. To find the final pressure and temperature, we must analyze this process using the laws of thermodynamics.

The Master Equation

Let's break this down thermodynamically. The walls of the container are perfectly insulating, which means no heat can enter or leave the system. Therefore, the heat exchanged is zero (

Q = 0

Furthermore, because the gas is expanding into a vacuum, it is not pushing against any external opposing force. There is no external pressure (

p_{e x t} = 0

), so the work done by the gas is also zero (

W = 0

According to the First Law of Thermodynamics, the change in internal energy

Δ U

is given by:

Δ U = Q - W

Since both

Q

and

W

are zero,

Δ U

is exactly zero! The internal energy of the gas remains perfectly constant.

Here is the crucial catch: for an ideal gas, the internal energy depends exclusively on its temperature. If the internal energy doesn't change, the temperature cannot change either. Therefore, the final temperature remains exactly

T

Final Calculation

So we have established that the temperature is constant (

T_{f} = T

). This means we can treat the initial and final states as if they were connected by an isothermal process. Let's bring in Boyle's Law, which states that for a fixed mass of an ideal gas at a constant temperature, the product of pressure and volume is constant:

p_{i} V_{i} = p_{f} V_{f}

Let's substitute our values. The initial volume was

V

. After the valve is opened, the gas occupies both equal halves of the container, so the final volume becomes

V + V = 2 V

. Plugging this into our equation, we get:

p \cdot V = p_{f} \cdot (2 V)

The

V

on both sides beautifully cancels out, leaving us with the final pressure:

p_{f} = \frac{p}{2}

So, the pressure halves, but the temperature remains exactly the same. Our final state is $(\frac{p}{2}, T)$ .

Common Traps & FAQs

Why is the work done zero? Doesn't the gas expand?

Work is done when a gas expands against an external opposing force (like a piston or atmospheric pressure). In a vacuum, there is no external pressure (

p_{e x t} = 0

), so the gas expands freely without doing any work.

Why doesn't the temperature drop when the gas expands?

For an ideal gas, there are no intermolecular forces. Since no work is done and no heat is exchanged, the internal energy remains constant. Because the internal energy of an ideal gas depends solely on its temperature, the temperature must also remain constant.

What if the gas was not ideal?

If it were a real gas, the molecules would have to overcome intermolecular attractive forces as they move apart. This internal work would consume some kinetic energy, leading to a slight drop in temperature, known as the Joule-Thomson effect.

Can we use

p_{1} V_{1}^{γ} = p_{2} V_{2}^{γ}

since it's an adiabatic process?

No! The equation

p V^{γ} = constant

is only valid for *reversible* adiabatic processes. Free expansion is a highly irreversible process, so you cannot use this relation. You must rely on the First Law of Thermodynamics directly.

Why does the volume exactly double?

The problem states the container is divided into two *equal* parts. Initially, the gas occupies one part (volume

V

). When the valve opens, it occupies both parts, so the total volume becomes

V + V = 2 V

Similar Questions

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An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_{1}$ and contains ideal gas at pressure $p_{1}$ and temperature $T_{1}$ . The other chamber has volume $V_{2}$ and contains ideal gas at pressure $p_{2}$ and temperature $T_{2}$ . If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

(A)

\frac{T _{1} T _{2} ( p _{1} V _{1} + p _{2} V _{2} )}{p _{1} V _{1} T _{2} + p _{2} V _{2} T _{1}}

(B)

\frac{p _{1} V _{1} T _{1} + p _{2} V _{2} T _{2}}{p _{1} V _{1} + p _{2} V _{2}}

(C)

\frac{p _{1} V _{1} T _{2} + p _{2} V _{2} T _{1}}{p _{1} V _{1} + p _{2} V _{2}}

(D)

\frac{T _{1} T _{2} ( p _{1} V _{1} + p _{2} V _{2} )}{p _{1} V _{1} T _{1} + p _{2} V _{2} T _{2}}

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A container of volume $1 m^{3}$ is divided into two equal parts by a partition. One part has an ideal gas at $300 K$ and the other part is vacuum. The whole system is thermally isolated from the surroundings. When the partition is removed, the gas expands to occupy the whole volume. Its temperature will now be ......

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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)

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(C)

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(D)

P_{2} V_{2}^{γ} = P_{1} V_{1}^{γ}

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A thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $γ$ . It is moving with speed $v$ and its suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by

(A)

\frac{( γ - 1 )}{2 γ R} M v^{2} K

(B)

\frac{γ M v ^{2}}{2 R} K

(C)

\frac{( γ - 1 )}{2 R} M v^{2} K

(D)

\frac{( γ - 1 )}{2 ( γ + 1 ) R} M v^{2} K

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As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition ( $P_{1}$ ) and a freely movable but thermally insulated piston ( $P_{2}$ ). The partition $P_{1}$ with thermal conductivity $K$ , cross sectional area $A$ and width $x$ divides the container into two sections, $S_{1}$ and $S_{2}$ , each containing one mole of a monoatomic gas. The piston $P_{2}$ moves freely such that the gas in $S_{2}$ is always at the atmospheric pressure. Initially, the difference between the temperatures of $S_{1}$ and $S_{2}$ is $Δ T_{0}$ . The time it takes for the temperature difference to become $\frac{Δ T _{0}}{2}$ is $n x R / K A$ , where $R$ is the universal gas constant. The value of $n$ is: [Given: $ln 2 \approx 0.7$ ]

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Comprehension Passage

In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monoatomic gas are

C_{V} = \frac{3}{2} R, C_{p} = \frac{5}{2} R

, and those for an ideal diatomic gas are

C_{V} = \frac{5}{2} R, C_{p} = \frac{7}{2} R

Question 1:

Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

An insulated box containing a monoatomic gas of molar mass $M$ moving with a speed $v_{0}$ is suddenly stopped. Find the increment in gas temperature as a result of stopping the box.

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An ideal gas has a specific heat at constant pressure $C_{p} = \frac{5 R}{2}$ . The gas is kept in a closed vessel of volume $0.0083 m^{3}$ , at a temperature of $300 K$ and a pressure of $1.6 \times 1 0^{6} N/m^{2}$ . An amount of $2.49 \times 1 0^{4} J$ of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas.

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The left and right compartments of a thermally isolated container of length $L$ are separated by a thermally conducting, movable piston of area $A$ . The left and right compartments are filled with $\frac{3}{2}$ and 1 moles of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant $k$ and natural length $\frac{2 L}{5}$ . In thermodynamic equilibrium, the piston is at a distance $\frac{L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $P = \frac{k L}{A} α$ , then the value of $α$ is _____

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LEVELJEE Advanced

Comprehension Passage

A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume,

C_{V} = 2 R

. Here, R is the gas constant. Initially, each side has a volume

V_{0}

and temperature

T_{0}

. The left side has an electric heater, which is turned on at very low power to transfer heat Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to

V_{0} /2

. Consequently, the gas temperatures on the left and the right sides become

T_{L}

and

T_{R}

, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

Question 1:

The value of $\frac{T _{R}}{T _{0}}$ is

(A)

2

(B)

3

(C)

2

(D)

3

Question 2:

The value of $\frac{Q}{R T _{0}}$ is

(A)

4 (22 + 1)

(B)

4 (22 - 1)

(C)

(52 + 1)

(D)

(52 - 1)

Select Answer:

Visualized Solution

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

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

.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; } Change in Internal Energy

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

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

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

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

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

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

Comprehension Passage

Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of T0​TR​​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of RT0​Q​ is

$I ni t ia l S e t u p$

$F r ee E x p an s i o n$

$F i r s t L a w o f T h er m o d y nami cs$

$C han g e in I n t er na l E n er g y$

$F ina l T e m p er a t u r e$

$B oy l e^{'} s L a w$

$F ina l P r ess u r e$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The value of $\frac{T _{R}}{T _{0}}$ is

The value of $\frac{Q}{R T _{0}}$ is