A vessel contains 100 litres of a liquid X.Heat is supplied to the liquid in such a fashion that heat given = change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is 1 atm and 202.6joules of heat were supplied, then [U= total internal energy]

Question

A vessel contains 100 litres of a liquid X.Heat is supplied to the liquid in such a fashion that heat given = change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is 1 atm and 202.6joules of heat were supplied, then [U= total internal energy] (A) Δ U =0, Δ H =0 (B) Δ U =+202.6 J, Δ H =+202.6 J (C) Δ U =-202.6 J, Δ H =-202.6 J (D) Δ U =0, Δ H =+ 202.6 J

Tardigrade Admin · Accepted Answer

Δ H = heat given. So, process is isobaric. w = -P textext(Δ V) = - 1×2 litre, atm = -202.6J Hence, q = 202.6 J = Δ H and Δ E = q + w = 0

Q. A vessel contains 100 litres of a liquid $X$ .Heat is supplied to the liquid in such a fashion that heat given = change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is 1 atm and 202.6joules of heat were supplied, then [ $U =$ total internal energy]

$Δ U = 0, Δ H = 0$

$Δ U = + 202.6 J, Δ H = + 202.6 J$

$Δ U = - 202.6 J, Δ H = - 202.6 J$

$Δ U = 0, Δ H = + 202.6 J$

$Δ H =$ heat given. So, process is isobaric.
$w = - P_{ext} (Δ V) = - 1 \times 2$ litre, atm $= - 202.6 J$
Hence, $q = 202.6 J = Δ H$ and $Δ E = q + w = 0$

Q. A vessel contains 100 litres of a liquid X.Heat is supplied to the liquid in such a fashion that heat given = change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is 1 atm and 202.6joules of heat were supplied, then [U= total internal energy]

ΔU=0,ΔH=0

ΔU=+202.6J,ΔH=+202.6J

ΔU=−202.6J,ΔH=−202.6J

ΔU=0,ΔH=+202.6J