Internal energy change for reactions at constant volume vs constant pressure The reaction of cyanamide, $\ce{NH2CN(s)},$ with dioxygen was carried
  out in a bomb calorimeter, and $∆U$ was found to be $\pu{–742.7 kJ mol-1}$
  at $\pu{298 K}.$ Calculate enthalpy change for the reaction at $\pu{298 K}.$


$$\ce{NH2CN(s) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l)}$$

$ΔH$ could be found from

$$ΔH = ΔU + ΔnRT$$

I want to know whether $ΔU$ found for constant volume conditions (bomb calorimeter) can be used in constant pressure conditions (while calculating $ΔH$ pressure is constant). In other words, is $ΔU$ same for both constant pressure and constant volume conditions, and why?

I searched the entire net and several standard physical chemistry texts but couldn't find any explanation.
The key point to keep in mind in this type of problem is that the energy of an ideal gas is only a function of temperature, not of volume or pressure$^\ast$. We also assume that the dependence of the energy of the condensed phases on p and V is negligible. Therefore the final pressure or volume is not going to affect $\Delta U$ for the reaction, provided n and T are constant. Since here the reaction refers to a conversion of a stoichiometric amount of reactants into products at the specified T, we therefore do not expect much of a change in $\Delta U$ with change in p or V. 

$^\ast$Note that the total differential for U can be written as 
$$\begin{align} dU &amp;= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV + C_VdT \end{align}$$ Therefore at constant T,
$$\begin{align} dU = &amp;= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV  \end{align}$$ 
The term in parentheses is zero for an ideal gas.