Which of the following relations are correct?
$(A)$ $\Delta U = q + p \Delta V$
$(B)$ $\Delta G = \Delta H - T \Delta S$
$(C)$ $\Delta S = \frac{q_{rev}}{T}$
$(D)$ $\Delta H = \Delta U - \Delta nRT$
Choose the most appropriate answer from the options given below:
$(A)$ $\Delta U = q + p \Delta V$
$(B)$ $\Delta G = \Delta H - T \Delta S$
$(C)$ $\Delta S = \frac{q_{rev}}{T}$
$(D)$ $\Delta H = \Delta U - \Delta nRT$
Choose the most appropriate answer from the options given below:
- A$C$ and $D$ only
- B$B$ and $C$ only
- C$A$ and $B$ only
- D$B$ and $D$ only
Explore More
Similar Questions
$11.0 \ L$ of an ideal gas at a constant external pressure of $5 \ atm$ is compressed isothermally to a final volume of $1 \ L$. The heat absorbed and work done,respectively,during this compression (in $L \ atm$) are:
The heat of neutralization of a strong acid and a strong alkali is $-57.0 \ kJ \ mol^{-1}$. The heat released when $0.5 \ mole$ of $HNO_3$ solution is mixed with $0.2 \ mole$ of $KOH$ is ....$kJ$
Medium
View SolutionTwo statements are given below.
Statement $I$: The reaction $Cr_2O_3 + 2 Al \rightarrow Al_2O_3 + 2 Cr$ $(\Delta G^{\ominus} = -421 \ kJ)$ is thermodynamically feasible.
Statement $II$: The above reaction occurs at room temperature.
The correct answer is
Statement $I$: The reaction $Cr_2O_3 + 2 Al \rightarrow Al_2O_3 + 2 Cr$ $(\Delta G^{\ominus} = -421 \ kJ)$ is thermodynamically feasible.
Statement $II$: The above reaction occurs at room temperature.
The correct answer is
An ideal gas in a thermally insulated vessel at internal pressure $P_1$,volume $V_1$,and absolute temperature $T_1$ expands irreversibly as shown in the following diagram. The final pressure,volume,and absolute temperature of the gas are $P_2$,$V_2$,and $T_2$ respectively. Which is the correct statement for the expansion?
Difficult
View SolutionAt $300 \ K$,$3.0 \ \text{moles}$ of an ideal gas at $3.0 \ \text{atm}$ pressure is compressed isothermally to one half of its volume by an external pressure of $6.0 \ \text{atm}$. The work done (in $kJ$) is. Given,$R=0.082 \ \text{L atm K}^{-1} \text{mol}^{-1}$ $(1 \ \text{L atm} = 101.3 \ \text{J})$.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo