The internal energy of an ideal gas increases during an isothermal process when the gas is

One mole of a non-ideal gas undergoes a change of state ($2.0 \, atm$,$3.0 \, L$,$95 \, K$) $\rightarrow$ ($4.0 \, atm$,$5.0 \, L$,$245 \, K$) with a change in internal energy,$\Delta U = 30.0 \, L \, atm$. The change in enthalpy $\Delta H$ of the process in $L \, atm$ is.

For an ideal gas,the heat of reaction at constant pressure and constant volume are related as:

$10$ moles of an ideal gas confined to a volume of $10 \, L$ is released into the atmosphere at $300 \, K$ where the pressure is $1 \, bar$. The work done by the gas is ....$L \, bar$
$(R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1})$

Consider the following statements:
$(1)$ Joule-Thomson experiment is isoenthalpic as well as adiabatic.
$(2)$ $A$ negative value of $\mu_{JT}$ (Joule-Thomson coefficient) corresponds to warming of a gas on expansion.
$(3)$ The temperature at which neither cooling nor heating effect is observed is known as inversion temperature.
Which of the above statements are correct?

For the reaction of nitrogen with hydrogen to form ammonia,$\Delta H^o = -92.2 \ kJ/mol$. If the reaction is carried out at a constant pressure of $20.0 \ atm$ and a volume change of $-1.16 \ L$,calculate the change in internal energy $(\Delta U)$ in $kJ$.