(C) $(P)$ $\rightarrow (3), (Q)$ $\rightarrow (4), (R)$ $\rightarrow (1), (S)$ $\rightarrow (2)$$I \rightarrow$ adiabatic$II \rightarrow$ isobaric$III

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$P$ $\rightarrow 3, Q$ $\rightarrow 4, R$ $\rightarrow 1, S$ $\rightarrow 2$

(C) $(P)$ $\rightarrow (3), (Q)$ $\rightarrow (4), (R)$ $\rightarrow (1), (S)$ $\rightarrow (2)$$I \rightarrow$ adiabatic$II \rightarrow$ isobaric$III \rightarrow$ isochoric$IV \rightarrow$ isothermal$(P)$ Process $I$ is adiabatic. Hence,no heat is exchanged between gas and surrounding.$(Q)$ Process $II$ is isobaric. The work done is $w = P \Delta V = 3 P_0 (3 V_0 - V_0) = 6 P_0 V_0$.$(R)$ Process $III$ is isochoric. Since volume is constant,$\Delta V = 0$,so $w = 0$.$(S)$ Process $IV$ is isothermal,where $T =$ constant.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

Medium

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the $PV$-diagram below. Among these four processes,one is isobaric,one is isochoric,one is isothermal and one is adiabatic. Match the processes mentioned in List-$I$ with the corresponding statements in List-$II$.

List-$I$ List-$II$

$P$. In process $I$ $1$. Work done by the gas is zero

$Q$. In process $II$ $2$. Temperature of the gas remains unchanged

$R$. In process $III$ $3$. No heat is exchanged between the gas and its surroundings

$S$. In process $IV$ $4$. Work done by the gas is $6 P_0 V_0$

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A
$P$ $\rightarrow 4, Q$ $\rightarrow 3, R$ $\rightarrow 1, S$ $\rightarrow 2$
B
$P$ $\rightarrow 1, Q$ $\rightarrow 3, R$ $\rightarrow 2, S$ $\rightarrow 4$
C
$P$ $\rightarrow 3, Q$ $\rightarrow 4, R$ $\rightarrow 1, S$ $\rightarrow 2$
D
$P$ $\rightarrow 3, Q$ $\rightarrow 4, R$ $\rightarrow 2, S$ $\rightarrow 1$

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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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The amount of heat evolved when $500 \ cm^{3}$ of $0.1 \ M \ HCl$ is mixed with $200 \ cm^{3}$ of $0.2 \ M \ NaOH$ is (in $kJ$)

DifficultKCET 2011

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At the temperature $T$ $(K)$ for the reaction: $X_2O_{4(l)} \rightarrow 2XO_{2(g)}$,given $\Delta U = x \ kJ \ mol^{-1}$ and $\Delta S = y \ J \ K^{-1} \ mol^{-1}$. The Gibbs energy change for the reaction is: (Assume $X_2O_4$ and $XO_2$ are ideal gases)

EasyAP EAMCET 2018

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$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$
$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$
Therefore:

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Change in entropy for an ideal gas in a reversible isothermal process is given by:

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List-$I$	List-$II$
$P$. In process $I$	$1$. Work done by the gas is zero
$Q$. In process $II$	$2$. Temperature of the gas remains unchanged
$R$. In process $III$	$3$. No heat is exchanged between the gas and its surroundings
$S$. In process $IV$	$4$. Work done by the gas is $6 P_0 V_0$

Similar Questions

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 $.5$)

The amount of heat evolved when $500 \ cm^{3}$ of $0.1 \ M \ HCl$ is mixed with $200 \ cm^{3}$ of $0.2 \ M \ NaOH$ is (in $kJ$)

At the temperature $T$ $(K)$ for the reaction: $X_2O_{4(l)} \rightarrow 2XO_{2(g)}$,given $\Delta U = x \ kJ \ mol^{-1}$ and $\Delta S = y \ J \ K^{-1} \ mol^{-1}$. The Gibbs energy change for the reaction is: (Assume $X_2O_4$ and $XO_2$ are ideal gases)

$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$Therefore:

Change in entropy for an ideal gas in a reversible isothermal process is given by:

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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 $.5$)

$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$
$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$
Therefore: