The difference between $\Delta H$ and $\Delta U$ $(\Delta H - \Delta U),$ when the combustion of one mole of heptane $(l)$ is carried out at a temperature $T$ is equal to (in $,RT$)

  • A
    $-4$
  • B
    $-3$
  • C
    $3$
  • D
    $4$

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The enthalpy of sublimation of $I_{2(s)}$ is $57.3 \, kJ \, mol^{-1}$ and the enthalpy of fusion is $15.5 \, kJ \, mol^{-1}$. The enthalpy of vaporization of $I_2$ is ..................... $kJ \, mol^{-1}$. (in $.8$)

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One mole of an ideal monoatomic gas undergoes two reversible processes ($A \rightarrow B$ and $B \rightarrow C$) as shown in the given figure:
$A \rightarrow B$ is an adiabatic process. If the total heat absorbed in the entire process ($A \rightarrow B$ and $B \rightarrow C$) is $R T_2 \ln 10$,the value of $2 \log V_3$ is . . . . . [Use,molar heat capacity of the gas at constant pressure,$C_{p, m} = \frac{5}{2} R$ ]

Benzene burns according to the following equation at $300 \ K$ $(R = 8.314 \ J \ mol^{-1} K^{-1})$: $2 C_6H_{6(l)} + 15 O_{2(g)} \to 12 CO_{2(g)} + 6 H_2O_{(l)}$,$\Delta H^o = -6542 \ kJ/mol$. What is the $\Delta E^o$ for the combustion of $1.5 \ mol$ of benzene in $kJ$?

For an ideal gas, consider only $P-V$ work in going from an initial state $X$ to the final state $Z$. The final state $Z$ can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [take $\Delta S$ as change in entropy and $w$ as work done].
(A) $\Delta S_{X \to Z} = \Delta S_{X \to Y} + \Delta S_{Y \to Z}$
(B) $w_{X \to Z} = w_{X \to Y} + w_{Y \to Z}$
(C) $w_{X \to Y \to Z} = w_{X \to Y} + w_{Y \to Z}$
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If $92 \ g$ of $Na$ reacts with water in an open vessel at $300 \ K$,what is the value of work done (in $J$)? $[$Assume ideal nature of the gaseous product$]$

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