The slopes of isothermal and adiabatic curves are related as

  • A
    Isothermal curve slope = adiabatic curve slope
  • B
    Isothermal curve slope = $\gamma \times $ adiabatic curve slope
  • C
    Adiabatic curve slope = $\gamma \times $ isothermal curve slope
  • D
    Adiabatic curve slope = $\frac{1}{2} \times $ isothermal curve slope

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Three moles of an ideal monoatomic gas perform a cycle as shown in the figure. The gas temperatures in different states are: $T_1 = 400\, K, T_2 = 800\, K, T_3 = 2400\, K$ and $T_4 = 1200\, K.$ The work done by the gas during the cycle is ........ $kJ$.

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Three moles of an ideal gas $\left( {{C_P} = \frac{7}{2}R} \right)$ at pressure ${P_A}$ and temperature ${T_A}$ is isothermally expanded to twice its initial volume. It is then compressed at constant pressure to its original volume. Finally, the gas is compressed at constant volume to its original pressure ${P_A}.$ The correct $P-V$ and $P-T$ diagrams indicating the process are

$A$ thermodynamic cycle of an ideal gas is shown in the figure. Choose the correct option which represents the same cycle.

The initial state of a certain gas is $(P_i, V_i, T_i)$. It undergoes expansion until its volume becomes $V_f$. Consider the following two cases:
$(a)$ The expansion takes place at constant temperature.
$(b)$ The expansion takes place at constant pressure.
Plot the $P-V$ diagram for each case. In which of the two cases is the work done by the gas more?

Two moles of monoatomic gas is expanded from $(P_0, V_0)$ to $(P_0, 2V_0)$ under isobaric condition. Let $\Delta Q_1$,$\Delta W_1$,and $\Delta U_1$ be the heat given to the gas,the work done by the gas,and the change in internal energy,respectively. Now,the monoatomic gas is replaced by a diatomic gas,with other conditions remaining the same. The corresponding values in this case are $\Delta Q_2$,$\Delta W_2$,and $\Delta U_2$. Then:

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