An ideal gas expands against a constant external pressure of $2 \ bar$ from $5 \ L$ to $8 \ L$ and absorbs $10 \ kJ$ of heat. What is $\Delta U$ of the system (in $J$)?

If $2 \ kJ$ of heat is released from the system and $6 \ kJ$ of work is done on the system,what is the internal energy change $(\Delta U)$ of the system?

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

Assertion : For an isothermal reversible process $Q = -W$,i.e.,work done by the system equals the heat absorbed by the system.
Reason : Enthalpy change $(\Delta H)$ is zero for an isothermal process.

Three moles of an ideal gas expand reversibly under isothermal conditions from $2 \, L$ to $20 \, L$ at $300 \, K$. The amount of heat change (in $kJ / mol$) in the process is:

For water $\Delta_{vap} H = 41 \ kJ \ mol^{-1}$ at $373 \ K$ and $1 \ bar$ pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water,the internal energy change during evaporation of water is $...... \ kJ \ mol^{-1}$.
[Use: $R = 8.3 \ J \ mol^{-1} \ K^{-1}$]