One mole of water is converted into steam at $373 \, K$. The heat absorbed at $1 \, atm$ pressure is $40.68 \, kJ$. If the molar volumes of water and steam are $18 \, mL$ and $30600 \, mL$ respectively,find $\Delta U$ for the process in $kJ$.

  • A
    $35.75$
  • B
    $31.75$
  • C
    $39.75$
  • D
    $37.60$

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An ideal gas is expanded from $(p_1, V_1, T_1)$ to $(p_2, V_2, T_2)$ under different conditions. The correct statement$(s)$ among the following is(are):
[$A$] The work done on the gas is maximum when it is compressed irreversibly from $(p_2, V_2)$ to $(p_1, V_1)$ against constant pressure $p_1$.
[$B$] The work done by the gas is less when it is expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions as compared to that when expanded reversibly from $V_1$ to $V_2$ under isothermal conditions.
[$C$] The change in internal energy of the gas is $(i)$ zero,if it is expanded reversibly with $T_1=T_2$,and $(ii)$ positive,if it is expanded reversibly under adiabatic conditions with $T_1 \neq T_2$.
[$D$] If the expansion is carried out freely,it is simultaneously both isothermal as well as adiabatic.

The difference between the reaction enthalpy change $(\Delta _r H)$ and reaction internal energy change $(\Delta _r U)$ for the reaction $2C_6H_{6(l)} + 15O_{2(g)} \longrightarrow 12CO_{2(g)} + 6H_2O_{(l)}$ at $300 \ K$ is $....$ $J \ mol^{-1}$ $(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

Match the following:
$A$. Isothermal process$i$. $q = \Delta U$
$B$. Adiabatic process$ii$. $W = - P \times \Delta V$
$C$. Isobaric process$iii$. $W = \Delta U$
$D$. Isochoric process$iv$. $W = - nRT \ln \left(\frac{v_f}{v_i}\right)$

The following reaction occurs in an automobile: $2C_8H_{18(g)} + 25O_{2(g)} \to 16CO_{2(g)} + 18H_2O_{(g)}$. The signs of $\Delta H$,$\Delta S$,and $\Delta G$ would be:

Expansion of $1$ $mol$ of an ideal gas takes place from $2$ $L$ to $8$ $L$ at $300$ $K$ against a constant external pressure of $1$ $atm$. Calculate $\Delta S_{total}$ in $J$ $K^{-1}$ $mol^{-1}$.
(Given: $R = 8.3$ $J$ $K^{-1}$ $mol^{-1}$,$1$ $L$ $atm = 100$ $J$,$\ln 2 = 0.693$) (in $.5$)

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