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Question
A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0°C). (a) Find the mass of the air in the container when thermal equilibrium is reached. (b) The container is now placed in another bath containing boiling water (100°C). Find the mass of air in the container. (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.
Use R = 8.3 J K-1 mol-1
Solution
(a) Here ,
`V_1= 5×10^(-5) "m"^3`
`P_1=10^5 "Pa"`
`T_1 = 273K`
M = 28.8 g
`P_1 V_1 = nRT_1`
⇒ `n =( P_1 V_1)/(RT_1)`
⇒ `m/M = (10^5×5×10^-5)/(8.3×273)`
⇒ `m = (10^5×5×10^-5×28.8)/(8.3×273)`
⇒ m = 0.0635 g
(b) Here,
`V_1 = 5×10^-5 "m"^3`
`P_1 = 10^5 "Pa"`
`P_2 = 10^5 "Pa"`
`T_1 = 273"K"`
`T_2 = 373"K"`
M = 28.8 g
`(P_1 V_1)/(T_1) = (P_2 V_2)/T_2`
`⇒ (5×10^-5)/273` = `V_2/373`
`⇒ V_2 = (5×10^-5×373)/273`
`⇒ V_2 = 6.831×10^-5`
Volume of expelled air `= 6.831×10^-5-5×10^-5`
=`1.831 ×10^-5`
Applying equation of state , we get
PV = nRT
`rArr m/M = (PV)/(RT)= (10^5×1.831×10^-5)/(8.3×373)`
`rArrm = (28.8×10^5×1.831×10^-5)/(8.3×373)=0.017`
Thus, mass of expelled air = 0.017 g
Amount of air in the container = 0.0635 - 0.017 = 0.0465 g
(c) Here,
T = 273K
`P = 10^5 `Pa
`V = 5×10^-5 "m"^3`
Applying equation of state, we get
PV = nRT
`⇒ P = (nRT)/"V" = (0.0465×8.3×273)/(28.8×5×10^-5)`
`P = 0.731×10^5 ≈ 73 "KPa"`
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