The first law of thermodynamics can be written as $\Delta U = \Delta Q + \Delta W$ for an ideal gas. Which of the following statements is correct?

  • A
    $\Delta U$ is always zero when no heat enters or leaves the gas.
  • B
    $\Delta W$ is the work done by the gas in this written law.
  • C
    $\Delta U$ is zero when heat is supplied and the temperature stays constant.
  • D
    $\Delta Q = -\Delta W$ when the temperature increases very slowly.

Explore More

Similar Questions

In a thermodynamic process,the pressure of a fixed mass of a gas is changed in such a manner that the gas releases $30 \,J$ of heat and $18 \,J$ of work is done on the gas. If the initial internal energy of the gas was $60 \,J$,what will be the final internal energy (in $\,J$)?

When the amount of work done is $333 \ cal$ and the change in internal energy is $167 \ cal$,then the heat supplied is ....... $cal$.

The internal energy change in a system that has absorbed $2 \, kcal$ of heat and done $500 \, J$ of work is ...... $J$.

Difficult
View Solution

$A$ sample of $0.1 \, g$ of water at $100^{\circ} C$ and normal pressure $(1.013 \times 10^5 \, N m^{-2})$ requires $54 \, cal$ of heat energy to convert to steam at $100^{\circ} C$. If the volume of the steam produced is $167.1 \, cc$,the change in internal energy of the sample is ....... $J$.

If $Q$,$E$,and $W$ denote respectively the heat added,change in internal energy,and the work done by a closed cycle process,then

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo