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 one atm, and 202.6 Joules of heat were supplied then, [U - total internal energy]

• A. $\Delta$U = 0 ,$\Delta$H = 0
• B. $\Delta$U = + 202. 6J, $\Delta$H = + 202.6 J
• C. $\Delta$U = – 202.6J, $\Delta$H = – 202.6J
• D. $\Delta$U = 0, $\Delta$H = + 202.6J

Answer 1

Answer: (D)

$\Delta$U = 0, $\Delta$H = + 202.6J

Answer 2

$\Delta$H = heat given. So, process is isobaric.

w = – Pext ($\Delta$V) = – 1 × 2 litre.atm = – 202.6 J

Hence, q = 202.6 J = $\Delta$H. and $\Delta$E = q + w = 0.