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

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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) $\to$ ( $4.0$ atm, $5.0$ L, $245$ K) with a change in internal energy $Δ U = 30.0$ L atm. The change in enthalpy 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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56.0

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Solution

The correct option is D $44.0$
Let us consider the problem:
$Δ H = H_{2} - H_{1} = U_{2} - U_{1} + P_{2} V_{2} - P_{1} V_{1}$
Hence,
$= 30 + 4 \times 5 - 2 \times 3 = 44 L . a t m .$

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One mole of a non ideal gas undergoes a change of state (2.0 atm, 3.0L, 95K)→(4.0 atm, 5.0L, 245 K) with a change in internal energy ΔU=30.0L atm. The change in enthalpy of the process in L atm is?

The correct option is D 44.0Let us consider the problem:ΔH=H2−H1=U2−U1+P2V2−P1V1Hence,=30+4×5−2×3=44L.atm.

One mole of a non ideal gas undergoes a change of state ( $2.0$ atm, $3.0$ L, $95$ K) $\to$ ( $4.0$ atm, $5.0$ L, $245$ K) with a change in internal energy $Δ U = 30.0$ L atm. The change in enthalpy of the process in L atm is?

The correct option is D $44.0$
Let us consider the problem:
$Δ H = H_{2} - H_{1} = U_{2} - U_{1} + P_{2} V_{2} - P_{1} V_{1}$
Hence,
$= 30 + 4 \times 5 - 2 \times 3 = 44 L . a t m .$