Question:
Find out the work done in the following process.
Find out the work done in the following process.
Updated On: May 19, 2024
- 61.5 Joule
- -61.5 Joule
- +246 Joule
- -246 Joule
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
The Correct answer is option is (A) : 61.5 Joule
Was this answer helpful?
0
1
Top Questions on Order of Reaction
- For a first-order reaction, the slope of the graph between \(\log[A]\) vs time is equal to:
- MHT CET - 2025
- Chemistry
- Order of Reaction
- A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled, (ii) reduced to half?
- CBSE CLASS XII - 2025
- Chemistry
- Order of Reaction
- For a first-order reaction, the concentration of a reactant decreases from 0.8 M to 0.4 M in 20 minutes. What is the half-life of the reaction? (Use $\ln 2 = 0.693$)
- AP EAPCET - 2025
- Chemistry
- Order of Reaction
The following data were obtained for the reaction: \[ 2NO(g) + O_2(g) \rightarrow 2N_2O(g) \] at different concentrations:
The rate law of this reaction is:- KEAM - 2025
- Chemistry
- Order of Reaction
- Which of the following plots gives a straight line for a zero-order reaction?
- MHT CET - 2025
- Chemistry
- Order of Reaction
View More Questions
Questions Asked in JEE Main exam
- 2.8 \( \times 10^{-3} \) mol of \( \text{CO}_2 \) is left after removing \( 10^{21} \) molecules from its ‘\( x \)’ mg sample. The mass of \( \text{CO}_2 \) taken initially is: Given: \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)
- JEE Main - 2025
- Mole concept and Molar Masses
- A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is ________________________.
- JEE Main - 2025
- Electromagnetic induction
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
- JEE Main - 2025
- Electromagnetic Field (EMF)
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
- JEE Main - 2025
- Thermodynamics
View More Questions