Question

A closed vessel contains pure water, in thermal equilibrium with its vapour at ${25}^{o}C$ (State #$1$). as shown



The vessel in this state is then kept inside an isothermal oven which is having an atmosphere of hot air maintained at ${80}^{o}C$. The vessel exchanges heat with the oven atmosphere and attains a new thermal equilibrium (Stage #$2$). If the Valve $A$ is now opened inside the oven, what will happen immediately after opening the valve?

• A. Nothing will happen - the vessel will continue to remain in equilibrium.
• B. Water vapor inside the vessel will come out other of the Valve A
• C. All the vapour inside the vessel will immediately condense
• D. Hot air will go inside the vessel through Value A

Answer 1

The correct option is D Hot air will go inside the vessel through Value A

Initially when water and water vapour mixture is at ${25}^{o}C$ then the maximum vapour pressure that can be at ${25}^{o}C$ in the saturation pressure of vapour at ${25}^{o}C$.

The saturation pressure at ${25}^{o}C$ is very less than $1atm$ $\left(101.3kPa\right)$. It is around $\mathrm{3..17}kPa$



Now when this vessel is kept in the oven at ${80}^{o}C$ then at ${80}^{o}C$ the saturation pressure of water is still less than $1$ atm. It is around $47.2kPa$.

The vapour pressure will reach $1atm$ when temperature is ${100}^{o}C$.

Hence at ${80}^{o}C$ also the pressure will be then $1$ atm $80$ when valve is opened air will enter the valve.