One mole of a monoatomic ideal gas goes through a thermodynamic cycle,as shown in the volume versus temperature $(V-T)$ diagram. The correct statement$(s)$ is/are :
[$R$ is the gas constant]
$(1)$ Work done in this thermodynamic cycle $(1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1)$ is $|W| = \frac{1}{2} RT_0$
$(2)$ The ratio of heat transfer during processes $1 \rightarrow 2$ and $2 \rightarrow 3$ is $\left|\frac{Q_{1 \rightarrow 2}}{Q_{2 \rightarrow 3}}\right| = \frac{5}{3}$
$(3)$ The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
$(4)$ The ratio of heat transfer during processes $1 \rightarrow 2$ and $3 \rightarrow 4$ is $\left|\frac{Q_{1 \rightarrow 2}}{Q_{3 \rightarrow 4}}\right| = \frac{1}{2}$

  • A
    $1, 3$
  • B
    $1, 2$
  • C
    $1, 4$
  • D
    $1, 3, 4$

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An ideal gas undergoes a cyclic process through various thermodynamic states. The heat energy $(Q)$ and work $(W)$ associated with these states are given as follows:
$Q_1 = 6000 \ J, Q_2 = -5500 \ J, Q_3 = -3300 \ J, Q_4 = 3500 \ J,$
$W_1 = 2500 \ J, W_2 = -1000 \ J, W_3 = -1200 \ J, W_4 = x \ J$
If the ratio of the net work done to the net heat absorbed is $\eta$, then the values of $x$ and $\eta$ are respectively:

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An ideal gas is subjected to a cyclic process involving four thermodynamic states. The amounts of heat $(Q)$ and work $(W)$ involved in each of these states are:
$Q_1 = 6000 \ J, Q_2 = -5500 \ J, Q_3 = -3000 \ J, Q_4 = 3500 \ J$
$W_1 = 2500 \ J, W_2 = -1000 \ J, W_3 = -1200 \ J, W_4 = x \ J$
The ratio of the net work done by the gas to the total heat absorbed by the gas is $\eta$. The values of $x$ and $\eta$ respectively are:

$A$ monoatomic gas at a pressure $P$,having a volume $V$ expands isothermally to a volume $4V$ and then adiabatically to volume $16V$. The final pressure of the gas is (Take $\gamma = \frac{5}{3}$)

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