One mole of a non-ideal gas undergoes a change of state 2-0 atm- 3-0 L- 95 K - 4-0 atm- 5-0 L- 245 K when a change in internal energy- U - 30-0 L atm- The change in enthalpy H of the process in L-atm is-

The correct option is C 44.0 H=U+PV As pressure is not constant, Δ H=H_2 H_1 #160; #160; #160; #160; =(U_2+P_2V_2) (U_1+P_1V_1) ...

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Standard XII

Chemistry

Heat of Reaction

One mole of a...

Question

One mole of a non-ideal gas undergoes a change of state ( $2.0$ atm, 3.0 L, 95 K) $\to$ ( $4.0 a t m, 5.0 L, 245 K$ ) when a change in internal energy, $△ U$ = 30.0 L atm. The change in enthalpy $(△ H)$ of the process (in L-atm) is:

40.0

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42.3

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44.0

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Not defined because pressure is not constant

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Solution

The correct option is C $44.0$
$H = U + P V$

As pressure is not constant,
$Δ H = H_{2} - H_{1}$

$= (U_{2} + P_{2} V_{2}) - (U_{1} + P_{1} V_{1})$

$= (U_{2} - U_{1}) + (P_{2} V_{2} - P_{1} V_{1})$

$= 30 + (4 \times 5 - 2 \times 3) = 44 L a t m$

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One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) → (4.0atm,5.0L,245K) when a change in internal energy, △U = 30.0 L atm. The change in enthalpy (△H) of the process (in L-atm) is:

The correct option is C 44.0H=U+PVAs pressure is not constant,ΔH=H2−H1 =(U2+P2V2)−(U1+P1V1) =(U2−U1)+(P2V2−P1V1) =30+(4×5−2×3)=44 L atm

One mole of a non-ideal gas undergoes a change of state ( $2.0$ atm, 3.0 L, 95 K) $\to$ ( $4.0 a t m, 5.0 L, 245 K$ ) when a change in internal energy, $△ U$ = 30.0 L atm. The change in enthalpy $(△ H)$ of the process (in L-atm) is:

The correct option is C $44.0$
$H = U + P V$

As pressure is not constant,
$Δ H = H_{2} - H_{1}$

$= (U_{2} + P_{2} V_{2}) - (U_{1} + P_{1} V_{1})$

$= (U_{2} - U_{1}) + (P_{2} V_{2} - P_{1} V_{1})$

$= 30 + (4 \times 5 - 2 \times 3) = 44 L a t m$