Question:
Following figure shows two processes A and B for a gas. If \(\Delta Q_A\) and \(\Delta Q_B\) are the amount of heat absorbed by the system in two cases, and \(\Delta U_A\) and \(\Delta U_B\) are changes in internal energies, respectively, then :
Following figure shows two processes A and B for a gas. If \(\Delta Q_A\) and \(\Delta Q_B\) are the amount of heat absorbed by the system in two cases, and \(\Delta U_A\) and \(\Delta U_B\) are changes in internal energies, respectively, then :
Show Hint
Internal energy is a state function (same initial and final states), while work is the area under the P-V curve.
Updated On: Nov 5, 2026
- \(\Delta Q_A > \Delta Q_B\) ; \(\Delta U_A > \Delta U_B\)
- \(\Delta Q_A \lt \Delta Q_B\) ; \(\Delta U_A \lt \Delta U_B\)
- \(\Delta Q_A > \Delta Q_B\) ; \(\Delta U_A = \Delta U_B\)
- \(\Delta Q_A = \Delta Q_B\) ; \(\Delta U_A = \Delta U_B\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The figure is a P-V diagram with two paths A and B from the same initial state i to the same final state f; path A lies above path B.
Internal energy is a state function, so \(\Delta U_A = \Delta U_B\).
Work done by the gas equals the area under the curve; path A (upper) encloses more area, so \(W_A > W_B\).
First law: \(\Delta Q = \Delta U + W\). Hence \(\Delta Q_A = \Delta U + W_A > \Delta Q_B = \Delta U + W_B\).
Answer: (3) \(\Delta Q_A > \Delta Q_B\) ; \(\Delta U_A = \Delta U_B\)
Internal energy is a state function, so \(\Delta U_A = \Delta U_B\).
Work done by the gas equals the area under the curve; path A (upper) encloses more area, so \(W_A > W_B\).
First law: \(\Delta Q = \Delta U + W\). Hence \(\Delta Q_A = \Delta U + W_A > \Delta Q_B = \Delta U + W_B\).
Answer: (3) \(\Delta Q_A > \Delta Q_B\) ; \(\Delta U_A = \Delta U_B\)
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- 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:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
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:
View More Questions
Top JEE Main Questions
- 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 ______.
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,
View More Questions