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Standard XII

Chemistry

Heat of Reaction

One mole of n...

Question

One mole of non-ideal gas undergoes a change of state $(1.0 a t m$ , $3.0 L$ , $200 K)$ to $(4.0 a t m, 5.0 L, 250 K)$ with a change in internal energy $(Δ U) = 40 L - a t m$ .
The change in enthalpy of the process in $L - a t m$ is :

43

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57

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42

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None of these

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Solution

The correct option is B $57$
Given, $(P_{1}, V_{1} = 1 a t m, 3 L)$
$(P_{2}, V_{2} = 4 a t m, 5 L)$

$Δ H = Δ U + P_{2} V_{2} - P_{1} V_{1}$

$= 40 + (4 \times 5) - (1 \times 3)$

$= 57$ L−atm
Option B is correct.

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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 non-ideal gas undergoes a change of state (1.0 atm, 3.0 L, 200 K) to (4.0 atm,5.0 L,250 K) with a change in internal energy (ΔU)=40 L−atm. The change in enthalpy of the process in L−atm is :

The correct option is B 57Given, (P1,V1=1 atm,3 L) (P2,V2=4 atm,5 L)ΔH=ΔU+P2V2–P1V1=40+(4×5)−(1×3)=57 L−atmOption B is correct.

One mole of non-ideal gas undergoes a change of state $(1.0 a t m$ , $3.0 L$ , $200 K)$ to $(4.0 a t m, 5.0 L, 250 K)$ with a change in internal energy $(Δ U) = 40 L - a t m$ .
The change in enthalpy of the process in $L - a t m$ is :

The correct option is B $57$
Given, $(P_{1}, V_{1} = 1 a t m, 3 L)$
$(P_{2}, V_{2} = 4 a t m, 5 L)$

$Δ H = Δ U + P_{2} V_{2} - P_{1} V_{1}$

$= 40 + (4 \times 5) - (1 \times 3)$

$= 57$ L−atm
Option B is correct.