A sample of diborane gas has a pressure of 345 torr at a temperature of -15c and volume of 3.48L if conditions are changed so that the temperature is 36c and the pressure is 468 torr what will be the volume of the sample

3.05L= V2

Explanation:

using

P1V1/T1=P2V2/T2

where p1= 345torr

T1, tempertaure= -15c+ 273=258K

T2, temperature= 36+ 273=309K

p2, Pressure= 468 torr

V2=???

P1V1/T1=P2V2/T2

v2=P1V1T2/T1P2= 345X3.46L X 309/ 258X 468=

V2=  3.05L

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

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enthalpy change for the reaction between the given volumes of the gases is :

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.