An ideal gas in a thermally insulated vessel | vedclass

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Question

(A) Since the vessel is thermally insulated,the heat exchange $q = 0$.Since the gas expands against zero external pressure $(P_{ex} = 0)$,the work don

Vedclass · Accepted Answer

$A, B, C$

Vedclass · Answer

(A) Since the vessel is thermally insulated,the heat exchange $q = 0$.Since the gas expands against zero external pressure $(P_{ex} = 0)$,the work done $w = -P_{ex} \Delta V = 0$.According to the first law of thermodynamics,$\Delta U = q + w = 0 + 0 = 0$.For an ideal gas,internal energy $U$ is a function of temperature only,so $\Delta U = 0$ implies $\Delta T = 0$,which means $T_2 = T_1$.Using the ideal gas equation $PV = nRT$,since $n, R,$ and $T$ are constant,$P_1 V_1 = P_2 V_2$.The process is an adiabatic irreversible expansion (free expansion),so the relation $P_2 V_2^\gamma = P_1 V_1^\gamma$ is not applicable.Therefore,statements $(A), (B),$ and $(C)$ are correct.

Column-$I$	Column-$II$
$(a)$ Adiabatic process	$(1)$ Heat
$(b)$ Isolated system	$(2)$ At constant volume
$(c)$ Isothermal change	$(3)$ First law of thermodynamics
$(d)$ Path function	$(4)$ No exchange of matter and energy
$(e)$ State function	$(5)$ No heat exchange
$(f)$ $\Delta U = q$	$(6)$ Constant temperature
$(g)$ Law of conservation of energy	$(7)$ Internal energy
$(h)$ Reversible process	$(8)$ $p_{ext} = 0$
$(i)$ Free expansion	$(9)$ At constant pressure
$(j)$ $\Delta H = q$	$(10)$ Infinitely slow process involving multiple equilibrium states
$(k)$ Intensive property	$(11)$ Entropy
$(l)$ Extensive property	$(12)$ Pressure,$(13)$ Specific heat

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