At $300 \ K$, $3.0 \ \text{moles}$ 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, $R=0.082 \ \text{L atm K}^{-1} \text{mol}^{-1}$ $(1 \ \text{L atm} = 101.3 \ \text{J})$.

  • A
    $7.476$
  • B
    $11.214$
  • C
    $3.738$
  • D
    $14.952$

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The combustion of one mole of benzene takes place at $298 \, K$ and $1 \, atm$. After combustion,$CO_{2(g)}$ and $H_2O_{(l)}$ are produced and $3267.0 \, kJ$ of heat is liberated. Calculate the standard enthalpy of formation,$\Delta_f H^{\ominus}$ of benzene. Standard enthalpies of formation of $CO_{2(g)}$ and $H_2O_{(l)}$ are $-393.5 \, kJ \, mol^{-1}$ and $-285.83 \, kJ \, mol^{-1}$ respectively.

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For a reaction $2 CO_{(g)} + O_{2(g)} \rightleftharpoons 2 CO_{2(g)}$,$\Delta_{r} G^0 = -128 \ kJ$ at $300 \ K$. If $\Delta_{r} S^0$ of the reaction is $-40 \ J \ K^{-1}$,calculate $\Delta_{r} U$ of the reaction. (in $kJ$)

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