For a pure substance, if $T_B$ is the melting point and $T_A$ is the freezing point, which graph correctly represents the relationship between the change in entropy $(\Delta S)$ and temperature $(T)$?

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    Option B
  • C
    Option C
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    Option D

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One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the $PV$-diagram below. Among these four processes,one is isobaric,one is isochoric,one is isothermal and one is adiabatic. Match the processes mentioned in List-$I$ with the corresponding statements in List-$II$.
List-$I$ List-$II$
$P$. In process $I$ $1$. Work done by the gas is zero
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The entropy versus temperature plot for phases $\alpha$ and $\beta$ at $1 \ bar$ pressure is given. $S_T$ and $S_0$ are entropies of the phases at temperatures $T$ and $0 \ K$,respectively.
The transition temperature for $\alpha$ to $\beta$ phase change is $600 \ K$ and $C_{p, \beta} - C_{p, \alpha} = 1 \ J \ mol^{-1} \ K^{-1}$. Assume $(C_{p, \beta} - C_{p, \alpha})$ is independent of temperature in the range of $200$ to $700 \ K$. $C_{p, \alpha}$ and $C_{p, \beta}$ are heat capacities of $\alpha$ and $\beta$ phases,respectively.
$(1)$ The value of entropy change,$S_{\beta} - S_{\alpha}$ (in $J \ mol^{-1} \ K^{-1}$),at $300 \ K$ is. . . . . . .
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For the reaction,$2 CO + O_2 \longrightarrow 2 CO_2$; $\Delta H = -560 \ kJ$. Two moles of $CO$ and one mole of $O_2$ are taken in a container of volume $1 \ L$. They completely form two moles of $CO_2$. The gases deviate appreciably from ideal behavior. If the pressure in the vessel changes from $70 \ atm$ to $40 \ atm$,find the magnitude (absolute value) of $\Delta U$ at $500 \ K$. $(1 \ L \ atm = 0.1 \ kJ)$

For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter,$\Delta U$ and $w$ correspond to

Calculate the enthalpy change when $50 \ mL$ of $0.01 \ M$ $Ca(OH)_2$ reacts with $25 \ mL$ of $0.01 \ M$ $HCl$. Given that $\Delta H^o$ for neutralization of a strong acid and a strong base is $-13.7 \ kcal \ mol^{-1}$. (Note: The provided value in the prompt $140 \ kcal \ mol^{-1}$ is physically incorrect for neutralization; using standard value $-13.7 \ kcal \ mol^{-1}$ for calculation). (in $kcal$)

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