$A$ monoatomic gas at pressure $P$ having volume $V$ expands isothermally to a volume $2V$ and then adiabatically to a volume $16V$. The final pressure of the gas is $\left(\gamma = \frac{5}{3}\right)$.

  • A
    $\frac{P}{64}$
  • B
    $\frac{P}{128}$
  • C
    $\frac{P}{8}$
  • D
    $\frac{P}{32}$

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Column-$I$ Column-$II$
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$A$ Carnot cycle for a gas is represented in the pressure-volume $(P-V)$ diagram shown below. Consider the following statements:
$I.$ The area of $ABCD$ is the work done by the gas.
$II.$ The area of $ABCD$ is the total heat absorbed.
$III.$ The change in internal energy of the cycle is zero.
Which of the following is correct?

Consider the thermodynamic cycle shown on the $PV$ diagram. The process $A \rightarrow B$ is isobaric,$B \rightarrow C$ is isochoric,and $C \rightarrow A$ is a straight-line process. The following internal energy change and heat are given: $\Delta U_{A \rightarrow B} = + 400 \text{ kJ}$ and $Q_{B \rightarrow C} = - 500 \text{ kJ}$. The heat flow in the process $Q_{C \rightarrow A}$ is ...... $\text{kJ}$.

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The figure shows the $P-V$ plot of an ideal gas taken through a cycle $ABCDA$. The part $ABC$ is a semi-circle and $CDA$ is half of an ellipse. Then,
$(A)$ the process during the path $A \rightarrow B$ is isothermal
$(B)$ heat flows out of the gas during the path $B \rightarrow C \rightarrow D$
$(C)$ work done during the path $A \rightarrow B \rightarrow C$ is zero
$(D)$ positive work is done by the gas in the cycle $ABCDA$

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