(C) Process $AB$ is isochoric (constant volume $V_1$), so the work done $W_{AB} = P \Delta V = 0$.Process $BC$ is isothermal (constant temperature $T_

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$0, R{T_2} \ln \left( \frac{V_2}{V_1} \right), R({T_1} - {T_2})$

(C) Process $AB$ is isochoric (constant volume $V_1$), so the work done $W_{AB} = P \Delta V = 0$.Process $BC$ is isothermal (constant temperature $T_2$). For $1 \, \text{mole}$ of an ideal gas, the work done is $W_{BC} = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{RT_2}{V} \, dV = RT_2 \ln \left( \frac{V_2}{V_1} \right)$.Process $CA$ is isobaric (constant pressure). The work done is $W_{CA} = P \Delta V = R \Delta T = R(T_1 - T_2)$.

Class 11 Physics Thermodynamics Heat, Work done and Internal Energy from Graph

Medium

$A$ cyclic process for $1 \, \text{mole}$ of an ideal gas is shown in the figure in the $V-T$ diagram. The work done in $AB$, $BC$, and $CA$ respectively is:

A
$0, R{T_2} \ln \left( \frac{V_1}{V_2} \right), R({T_1} - {T_2})$
B
$R({T_1} - {T_2}), 0, R{T_1} \ln \frac{V_1}{V_2}$
C
$0, R{T_2} \ln \left( \frac{V_2}{V_1} \right), R({T_1} - {T_2})$
D
$0, R{T_2} \ln \left( \frac{V_2}{V_1} \right), R({T_2} - {T_1})$

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Heat, Work done and Internal Energy from Graph

One mole of an ideal gas is taken from $a$ to $b$ along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is $w_s$ and that along the dotted line path is $w_d$,then the integer closest to the ratio $w_d / w_s$ is

IIT 2010

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The heat energy absorbed by a system in going through a cyclic process shown in the figure is:

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As shown in the figure,the amount of heat absorbed along the path $ABC$ is $90\,J$ and the amount of work done by the system is $30\,J$. If the amount of work done along the path $ADC$ is $20\,J$,the amount of heat absorbed will be .... $J$.

Medium

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One mole of an ideal gas in initial state $A$ undergoes a cyclic process $ABCA$,as shown in the figure. Its pressure at $A$ is $P_0$. Choose the correct option$(s)$ from the following:
$(A)$ Internal energies at $A$ and $B$ are the same
$(B)$ Work done by the gas in process $AB$ is $P_0 V_0 \ln 4$
$(C)$ Pressure at $C$ is $\frac{P_0}{4}$
$(D)$ Temperature at $C$ is $\frac{T_0}{4}$

DifficultIIT 2010

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An ideal gas undergoes a circular cycle centered at $4 \, atm, 4 \, L$ as shown in the diagram. The maximum temperature attained in this process is close to

KVPY 2015

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$A$ cyclic process for $1 \, \text{mole}$ of an ideal gas is shown in the figure in the $V-T$ diagram. The work done in $AB$, $BC$, and $CA$ respectively is:

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One mole of an ideal gas is taken from $a$ to $b$ along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is $w_s$ and that along the dotted line path is $w_d$,then the integer closest to the ratio $w_d / w_s$ is

The heat energy absorbed by a system in going through a cyclic process shown in the figure is:

As shown in the figure,the amount of heat absorbed along the path $ABC$ is $90\,J$ and the amount of work done by the system is $30\,J$. If the amount of work done along the path $ADC$ is $20\,J$,the amount of heat absorbed will be .... $J$.

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