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**1. Homework Statement**

The internal energy of air is given, at ordinary temperatures, by

## u= u_o + 0.718t ##

where ##u## is in kJ/kg, ##u_o## is any arbitrary value of ##u## at 0

^{o}C, kJ/kg, and t is temperature in

^{o}C.

Also for air,

##pv= 0.287(t+273)##

where p is in kPa and ##v## is in m

^{3}/kg.

(a) An evacuated bottle is fitted with a valve through which air from the atmosphere, at 760mm of Hg and 25

^{o}C, is allowed to flow slowly to fill the bottle. If no heat is transferred to or from the air in the bottle, what will it's temperature be when the pressure in the bottle reaches 760mm of Hg ?

Ans 144.2

^{o}C

(b) If the bottle initially contains 0.03 m

^{3}of air at 400mm Hg and 25

^{o}C, what will the temperature be when the pressure in the bottle reaches 760mm Hg ?

Ans 71.6

^{o}C

**2. Homework Equations**

## u= u_o + 0.718t ##

##pv= 0.287(t+273)##

##PV= mRT ##

##Q-W= \Delta u ##

Q- heat transfer

W- work transfer

##\Delta u ## = Internal energy change.

**3. The Attempt at a Solution**

(a)

##P_1 = 13600*9.81*0.760 = 101396.16 Pa##

##T_1## = 25

^{o}C

##u_1 = u_o +0.718(25) = u_o +17.95 kJ/kg ##

To find ##T_2##

##Q-W= \Delta u ##

Q= 0 as no heat is transferred , W=0 no work is done in free expansion ??

So ##\Delta u## has to be zero and T

_{1}=T

_{2}??

I don't know how they got an increase in temperature.

(b)

##V_1 = 0.03 m^3 ##

##P_1 = 13600*9.81*0.400 = 53366.4 Pa##

##T_1 = 298K ##

##P_2= 13600*9.81*0.760 = 101396.16 Pa##

To find ##T_2 ##

Since volume of the bottle is constant ##V_1= V_2## ??

Initial mass = ##m_1 = \frac{P_1 V_1} { RT_1} = \frac{53366.4*0.03} {287*298} = 0.01871936 ## kg

Final mass = ##m_2 =\frac {P_2 V_2} {RT_2} = \frac{101396.16*0.03} {287*T_2} = \frac {10.5890} {T_2} ## kg ...is this useful ???

Can i use ## \frac{P_1 V_1} {T_1} = \frac{ P_2 V_2} {T_2} ## ??

I get T_2 = 566.2 K ......clearly wrong , how to continue ?