$A$ system consisting of $1 \, mole$ of an ideal gas undergoes a reversible process,$A$ $\rightarrow B$ $\rightarrow C$ $\rightarrow A$ (schematically indicated in the figure below). If the temperature at the starting point $A$ is $300 \, K$ and the work done in the process $B \rightarrow C$ is $1 \, L \, atm$,the heat exchanged in the entire process in $L \, atm$ is $....$

  • A
    $1.0$
  • B
    $0.0$
  • C
    $1.5$
  • D
    $0.5$

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Similar Questions

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. . . . . . .
$(2)$ The value of enthalpy change,$H_{\beta} - H_{\alpha}$ (in $J \ mol^{-1}$),at $300 \ K$ is.
[Use : $\ln 2 = 0.69$,Given : $S_{\beta} - S_{\alpha} = 0$ at $0 \ K$]

For the reaction
$A_{(\ell)} \rightarrow 2 B_{(g)}$
$\Delta U = 2.1 \; kcal, \Delta S = 20 \; cal \; K^{-1} \; mol^{-1}$ at $300 \; K$
Hence $\Delta G$ in $kcal \; mol^{-1}$ is

$20.0 \ dm^{3}$ of an ideal gas '$X$' at $600 \ K$ and $0.5 \ MPa$ undergoes isothermal reversible expansion until the pressure of the gas is $0.2 \ MPa$. Which of the following options is correct? (Given: $\log\,2=0.3010$ and $\log\,5=0.6989$)

Consider the reaction at $300 \ K$: $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$; $\Delta H^o = -185 \ kJ$. If $2 \ mole$ of $H_2$ completely react with $2 \ mole$ of $Cl_2$ to form $HCl$,what is $\Delta U^o$ for this reaction 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) $w_{X \to Y \to Z} = w_{X \to Y} + w_{Y \to Z}$
(D) $\Delta S_{X \to Y \to Z} = \Delta S_{X \to Y}$

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