A gas has an internal energy given by the relation U = 1.5 pV . The...

The process is isobaric, so the pressure $p$ of the gas is constant and equal to the external pressure.

The work done by the gas during the expansion is given by $W = p \Delta V = p(V_2 - V_1)$.

Substituting the given values:

$W = 1.0 \times 10^5 \times (200 - 100) \times 10^{-6} = 1.0 \times 10^5 \times 10^{-4} = 10 \text{ J}$

The change in internal energy is $\Delta U = U_2 - U_1 = 1.5 p_2 V_2 - 1.5 p_1 V_1$.

Since the pressure is constant, $p_1 = p_2 = p$, giving $\Delta U = 1.5 p(V_2 - V_1) = 1.5 W$.

$\Delta U = 1.5 \times 10 = 15 \text{ J}$

According to the first law of thermodynamics, the heat absorbed by the gas is $\Delta Q = \Delta U + W$.

$\Delta Q = 15 + 10 = 25 \text{ J}$