One mole of diatomic ideal gas undergoes a cy | vedclass

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(B) For a diatomic ideal gas,the molar heat capacity at constant volume is $C_V = \frac{5}{2}R$.The change in internal energy $\Delta U$ for any proce

Vedclass · Accepted Answer

The change in internal energy in the process $BC$ is $-500\,R$.

Vedclass · Answer

(B) For a diatomic ideal gas,the molar heat capacity at constant volume is $C_V = \frac{5}{2}R$.The change in internal energy $\Delta U$ for any process is given by $\Delta U = n C_V \Delta T$.$1$. For process $AB$ (isochoric): $\Delta U_{AB} = n C_V (T_B - T_A) = 1 	imes \frac{5}{2}R 	imes (800 - 400) = 1000\,R$.$2$. For process $BC$ (adiabatic): $\Delta U_{BC} = n C_V (T_C - T_B) = 1 	imes \frac{5}{2}R 	imes (600 - 800) = -500\,R$.$3$. For process $CA$ (isobaric): $\Delta U_{CA} = n C_V (T_A - T_C) = 1 	imes \frac{5}{2}R 	imes (400 - 600) = -500\,R$.$4$. For the whole cyclic process,the change in internal energy is $\Delta U_{total} = \Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 1000\,R - 500\,R - 500\,R = 0$.Comparing these results with the given options,option $B$ is correct.

Curve	Process
$I$	$a)$ Adiabatic
$II$	$b)$ Isobaric
$III$	$c)$ Isochoric
$IV$	$d)$ Isothermal

Column $I$	Column $II$
$(A)$ Process $A \rightarrow B$	$(p)$ Internal energy decreases.
$(B)$ Process $B \rightarrow C$	$(q)$ Internal energy increases.
$(C)$ Process $C \rightarrow D$	$(r)$ Heat is lost.
$(D)$ Process $D \rightarrow A$	$(s)$ Heat is gained.
	$(t)$ Work is done on the gas.

One mole of diatomic ideal gas undergoes a cyclic process $ABC$ as shown in the figure. The process $BC$ is adiabatic. The temperatures at $A, B$,and $C$ are $400\,K, 800\,K$,and $600\,K$ respectively. Choose the correct statement.

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