One mole of a non-ideal gas undergoes a change of state :
( $2.0$ atm, $3.0$ L, $95 (K) ⟶ (4.0 a t m, 5.0 L, 245 K)$ with 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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None of the above

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Solution

The correct option is D $44.0$
Solution:- (C) $44.0$
Given:-
$(2 atm, 3.0 L, 95 K) \to (4 atm, 5.0 L, 245 K)$

Therefore,
$P_{1} = 2 a t m, P_{2} = 4 a t m$
$V_{1} = 3 L, V_{2} = 5 L$
$T_{1} = 95 K T_{2} = 245 K$
$Δ U = 30 L . a t m$

As we know that,
$Δ H = Δ U + (P_{2} V_{2} - P_{1} V_{1})$
$∴ Δ H = 30 + (4 \times 5 - 2 \times 3)$
$\Rightarrow Δ H = 30 + 14 = 44 L . a t m$

Hence the change in enthalpy is $44 L . a t m$ .

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Similar questions

Q. 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:

Q. One mole of a non- ideal gas undergoes a change of state

(2.0 a t m, 3.0 L, 95 K) \to (4.0 a t m, 5.0 L, 245 K)

with a change in internal energy,

△ U = 30.0 L

atm. The change in enthalpy

(△ H)

of the process is:

One mole of an ideal gas undergoes a change of state (2.0atm, 30L, 95K) $\to$ (4.0atm, 5.0L, 245K) with change in internal energy $Δ E = 30.0 L a t m$ .The change in enthalpy of the process in L atm is

One mole of a non - ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95K) $\to$ (4.0 atm, 5.0L, 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 ?

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

The correct option is D 44.0Solution:- (C) 44.0Given:-(2 atm ,3.0 L ,95 K)→(4 atm ,5.0 L ,245 K)Therefore,P1=2atm,P2=4atmV1=3L,V2=5LT1=95KT2=245KΔU=30L.atmAs we know that,ΔH=ΔU+(P2V2−P1V1)∴ΔH=30+(4×5−2×3)⇒ΔH=30+14=44L.atmHence the change in enthalpy is 44L.atm.

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