A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiabatic constant for the gas is \(\gamma=\frac{3}{2}\) , then the work done by the gas in the process is:
- \(RT[2-\sqrt2]\)
- \(\frac{R}{T}[2-\sqrt2]\)
- \(RT[2+\sqrt2]\)
- \(\frac{T}{R}[2-\sqrt2]\)
The Correct Option is A
Approach Solution - 1
To find the work done by the gas during adiabatic expansion, we will use the following principle of adiabatic processes: For adiabatic processes, the formula that relates initial and final states is given by:
\[ PV^\gamma = \text{constant} \]
where \( \gamma \) is the adiabatic constant, here being \( \frac{3}{2} \).
We will consider the initial and final states and the principle that in adiabatic processes, \( PV^\gamma = \text{constant} \).
Suppose the initial volume is \( V \) and the initial pressure is \( P \). After expansion, the final volume is \( 2V \) (since the volume is doubled). Let the final pressure be \( P' \).
By using the adiabatic condition:
\[ P \cdot V^\gamma = P' \cdot (2V)^\gamma \]
Simplifying further,
\[ P = P' \cdot 2^\gamma \]
Substitute \( \gamma = \frac{3}{2} \):
\[ P = P' \cdot 2^{\frac{3}{2}} \]
The work done \( W \) in an adiabatic process is given by:
\[ W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \]
Substituting \( \gamma \) and rearranging terms, we get:
\[ W = \frac{P \cdot V - P \cdot 2^{-\frac{1}{2}} \cdot 2V}{\frac{3}{2} - 1} \]
Simplifying the expression:
\[ W = \frac{P \cdot V (1 - \sqrt{2})}{\frac{1}{2}} = 2P \cdot V (1 - \sqrt{2}) \]
Using the ideal gas law, \( PV = nRT \), where \( n = 1 \) mole:
\[ W = 2 \cdot (RT) \cdot (1 - \sqrt{2}) \]
Thus, the work done by the gas in the process is:
\[ W = RT[2 - \sqrt{2}] \]
Therefore, the correct option is: \(RT[2 - \sqrt{2}]\).
Approach Solution -2
Step 1: Adiabatic condition The adiabatic condition is given by:
\(TV^{\gamma-1} = \text{constant.}\)
For the initial and final states:
\(T(V)^{\frac{3}{2}-1} = T_f(2V)^{\frac{3}{2}-1}.\)
Simplify:
\(TV^{\frac{1}{2}} = T_f(2V)^{\frac{1}{2}},\)
\(TV\sqrt{V} = T_f\sqrt{2V}.\)
Cancel \(\sqrt{V}\):
\(T = T_f\sqrt{2}.\)
Solve for \(T_f\):
\(T_f = \frac{T}{\sqrt{2}}.\)
Step 2: Work done in adiabatic expansion The work done in an adiabatic process is given by:
\(W.D. = \frac{nR}{1-\gamma}\left[T_f - T\right].\)
Substitute \(T_f = \frac{T}{\sqrt{2}}, \gamma = \frac{3}{2},\) and \(n = 1\):
\(W.D. = \frac{R}{1-\frac{3}{2}}\left[\frac{T}{\sqrt{2}} - T\right].\)
Simplify:
\(W.D. = \frac{R}{-\frac{1}{2}}\left[\frac{T}{\sqrt{2}} - T\right],\)
\(W.D. = -2R\left[\frac{T}{\sqrt{2}} - T\right].\)
Factorize:
\(W.D. = 2RT\left[1 - \frac{1}{\sqrt{2}}\right].\)
Simplify further:
\(W.D. = RT\left[2 - \sqrt{2}\right].\)
Final Answer: \(RT\left[2 - \sqrt{2}\right].\)
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Thermodynamics Questions
- Two thermodynamical processes are shown in the figure. The molar heat capacity for process \( A \) and \( B \) are \( C_A \) and \( C_B \). The molar heat capacity at constant pressure and constant volume are represented by \( C_P \) and \( C_V \), respectively. Choose the correct statement.
- During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of \(\frac{C_p}{C_v}\) for the gas is :
- A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. It's volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be :
- Choose the correct statement for processes A & B shown in figure.
- The pressure and volume of an ideal gas are related as \( PV^{3/2} = K \) (constant). The work done when the gas is taken from state A (\( P_1, V_1, T_1 \)) to state B (\( P_2, V_2, T_2 \)) is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.