An ideal gas expands from volume V_1 to V_2. | vedclass

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(B) For an ideal gas,the change in internal energy is given by $\Delta U = nC_v\Delta T$. Since $C_v > 0$,$\Delta U$ is directly proportional to the c

Vedclass · Accepted Answer

$\Delta U$ is greatest under adiabatic process.

Vedclass · Answer

(B) For an ideal gas,the change in internal energy is given by $\Delta U = nC_v\Delta T$. Since $C_v > 0$,$\Delta U$ is directly proportional to the change in temperature $\Delta T$.$1$. For an isothermal process,the temperature remains constant,so $\Delta T = 0$,which implies $\Delta U = 0$.$2$. For an adiabatic process,the gas expands $(V_2 > V_1)$,so the gas does positive work $(W > 0)$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since $\Delta Q = 0$,we have $\Delta U = -W$. Thus,$\Delta U $3$. For an isobaric process,the pressure is constant. From the ideal gas law $PV = nRT$,since $V$ increases,$T$ must increase $(\Delta T > 0)$. Therefore,$\Delta U > 0$.Comparing these values: $\Delta U_{	ext{isobaric}} > 0$,$\Delta U_{	ext{isothermal}} = 0$,and $\Delta U_{	ext{adiabatic}} Thus,the order is $\Delta U_{	ext{isobaric}} > \Delta U_{	ext{isothermal}} > \Delta U_{	ext{adiabatic}}$.Statement $b$ claims $\Delta U$ is greatest under the adiabatic process,which is false.

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.

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