A gas expands from $2 L$ to $6 L$ against a constant pressure of $0.5$ atm on absorbing $200 J$ of heat Calculate the change in internal energy.

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Solution

$q = 200 J; Δ V = 6 - 2 = 4 l i t = 0.004 m^{3}$ ; $P = 0.5 \times 101325 = 50662.5 P a$

$Δ H = Δ U + P Δ V$ (enthalpy change)

At const, pressure (for isobar process ) $Δ H = q$

Hence, $Δ U = q - P Δ V$ ( $1^{s t}$ law of theromadynamics )

$Δ U = 200 - 50662.5 \times 0.004 = 200 - 202.65$

$Δ U = - 2.65 J$

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A gas expands from 2 L to 6 L against a constant pressure of 0.5 atm on absorbing 200 J of heat Calculate the change in internal energy.

q=200 J;ΔV=6−2=4 lit=0.004 m3 ; P=0.5×101325=50662.5 Pa ΔH=ΔU+PΔV (enthalpy change)At const, pressure (for isobar process ) ΔH=qHence, ΔU=q−PΔV (1st law of theromadynamics )ΔU=200−50662.5×0.004=200−202.65ΔU=−2.65 J

A gas expands from $2 L$ to $6 L$ against a constant pressure of $0.5$ atm on absorbing $200 J$ of heat Calculate the change in internal energy.