# at 27 ? C and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in...

## Question:

Air at 27{eq}^{\circ} {/eq}C and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligible, the exit temperature of air will be

(a) 10 {eq}^{\circ} {/eq}C

(b) 15 {eq}^{\circ} {/eq}C

(c) 20 {eq}^{\circ} {/eq}C

(d) 23 {eq}^{\circ} {/eq}C

(e) 27 {eq}^{\circ} {/eq}C

## Kinetic energy:

As we know that the energy which is neither created nor destroyed but it will interchange from one form to another form. The kinetic energy is the type of energy that exists in a system that is in motion with respect to the reference point.

## Answer and Explanation:

We have given that,

Inlet temperature {eq}\left( {T_1 } \right) = 27^0 C {/eq}

Pressure {eq}\left( P \right) = 1\;{\rm{atm}} {/eq}

We have to calculate the exit temperature {eq}\left( {T_2 } \right) {/eq}

As we know, In case of an ideal gas

The temperature of the system will remain constant when the system is going through a throttling process.

So, {eq}T_1 = T_2 = 27^0 C {/eq}

Hence, {eq}\left( e \right) {/eq} is our correct option.