(A) The correct matches are as follows:$(A)$ At constant volume,the heat exchanged is equal to the change in internal energy: $q_V = \Delta U$ $(III)$

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(A) The correct matches are as follows:$(A)$ At constant volume,the heat exchanged is equal to the change in internal energy: $q_V = \Delta U$ $(III)$.$(B)$ For an isothermal irreversible process,the work done is given by: $W = -p_{ex} (V_f - V_i)$ $(IV)$.$(C)$ For an isothermal reversible process,the work done is given by: $W = -2.303 nRT \log \frac{V_f}{V_i}$ $(I)$.$(D)$ For an adiabatic process,$q = 0$,so according to the first law of thermodynamics $(\Delta U = q + W)$,we have $W_{adiabatic} = \Delta U$ $(II)$.Therefore,the correct sequence is $A-III, B-IV, C-I, D-II$.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

Medium

Match the following items in List-$I$ with their corresponding expressions in List-$II$.
List-$I$ List-$II$
$A$. At constant volume the change in internal energy of a system $I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$
$B$. Isothermal irreversible change $II$. $W_{adiabatic} = \Delta U$
$C$. Isothermal reversible change $III$. $q_V = \Delta U$
$D$. Adiabatic change $IV$. $W = -p_{ex} (V_f - V_i)$
$V$. $\Delta U = \Delta H - \Delta nRT$

AP EAMCET 2018 All AP EAMCET

A
$A-III, B-IV, C-I, D-II$
B
$A-IV, B-I, C-III, D-V$
C
$A-III, B-IV, C-I, D-V$
D
$A-III, B-V, C-I, D-II$

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Mix Examples-Thermodynamics and Thermochemistry

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Calculate $\Delta H$ when $2 \ moles$ of solid benzoic acid undergo complete combustion at $300 \ K$ if $C_6H_5COOH_{(s)} + \frac{15}{2} O_{2(g)} \rightarrow 7CO_{2(g)} + 3H_2O_{(l)}$,$\Delta U_{reaction} = -750 \ kJ/mole$ [$R = 8 \ J/mole \cdot K$].

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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 $kJ$?

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Water vapor is an ideal gas. Calculate the internal energy change $(\Delta U)$ when $1 \, \text{mol}$ of water is vaporized at $1 \, \text{bar}$ pressure and $100 \, ^{\circ}C$. (Given: Molar enthalpy of vaporization of water at $1 \, \text{bar}$ and $373 \, K = 41 \, kJ \, \text{mol}^{-1}$ and $R = 8.3 \, J \, \text{mol}^{-1} \, K^{-1}$)

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Which of the following relations between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is correct for the given reactions?

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Which of the following is incorrect?

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List-$I$	List-$II$
$A$. At constant volume the change in internal energy of a system	$I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$
$B$. Isothermal irreversible change	$II$. $W_{adiabatic} = \Delta U$
$C$. Isothermal reversible change	$III$. $q_V = \Delta U$
$D$. Adiabatic change	$IV$. $W = -p_{ex} (V_f - V_i)$
	$V$. $\Delta U = \Delta H - \Delta nRT$

Similar Questions

Calculate $\Delta H$ when $2 \ moles$ of solid benzoic acid undergo complete combustion at $300 \ K$ if $C_6H_5COOH_{(s)} + \frac{15}{2} O_{2(g)} \rightarrow 7CO_{2(g)} + 3H_2O_{(l)}$,$\Delta U_{reaction} = -750 \ kJ/mole$ [$R = 8 \ J/mole \cdot K$].

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 $kJ$?

Which of the following relations between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is correct for the given reactions?

Which of the following is incorrect?

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