Two closed vessels of equal volume contains air at 105 kpa 300 k and are
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Answer:
In this case, let the initial pressure = P and final pressure = P'. Since (P/T) = constant, considering the system of two spheres of equal volume, we get,
300
P
+
300
P
=
300
P
′
+
400
P
′
here, P = 105k Pa
Solving we get, P
′
=120kPa
Explanation:
Two closed vessels of equal volune contain air at 105kPa, 300K and are connected through a narrow tube. If one of the vessels is mow maintained at 300K and the other at 400K, what will be the pressure in the vessels? or, p'=(2p_(0)T')/(T'+T_(0))=(2xx105kPaxx400K)/(400K+300K)=120kPa`.
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