Mixing of non-reacting gases is generally accompanied by

  • A
    Decrease in entropy
  • B
    Increase in entropy
  • C
    Change in enthalpy
  • D
    Change in free energy

Explore More

Similar Questions

Standard entropies of $X_2, Y_2$ and $XY_5$ are $70, 50$ and $110 \ J \ K^{-1} \ mol^{-1}$ respectively. The temperature in Kelvin at which the reaction $\frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5$ with $\Delta H = -35 \ kJ \ mol^{-1}$ will be at equilibrium is . . . . . . (Nearest integer).

One mole of an ideal diatomic gas $(C_V = 5 \ cal)$ was transformed from initial $25 \ ^{\circ}C$ and $1 \ L$ to the state when temperature is $100 \ ^{\circ}C$ and volume $10 \ L$. The entropy change of the process can be expressed as $(R = 2 \ cal / mol \cdot K)$ :-

Difficult
View Solution

When $2 \ mol$ of an ideal gas $\left( C_{p,m} = \frac{5}{2} R \right)$ is heated from $300 \ K$ to $600 \ K$ at constant pressure,the change in entropy of the gas $\left( \Delta S \right)$ is:

One mole of water at $100\,^{\circ}C$ is converted into steam at $100\,^{\circ}C$ at a constant pressure of $1\,atm$. The change in entropy is [heat of vaporisation of water at $100\,^{\circ}C = 540\,cal/g$].

What is the unit of entropy?

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