What pressure is exerted by a #8,60mol# quantity of gas at a temperature of #320K#, that is confined to a #16.0*L# volume?

Question

936 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-01T10:07:26

#1430.678# Pa

Using ideal gas formula: #PV=nRT# .
Here, #V = 16L = 16 dm^3#
#n = 8.6 "moles"#
#R = 8.314#
#T = 47+273.15= 320.15" ""Kelvin"#

Hence #P = (nRT)/V = (8.6*8.314*320.15)/(16)#
#=1430.678 Pa#

Answer 2 · 2017-10-01T10:15:49

Approx. #14*atm#.....

We use the Ideal Gas equation.....

#P=(nRT)/V=(8.60*cancel(mol)xx0.0821*(cancelL*atm)/(cancel(K)*cancel(mol))xx320*cancelK)/(16.0*cancelL)#

#=??*atm#....

What pressure is exerted by a #8,60*mol# quantity of gas at a temperature of #320*K#, that is confined to a #16.0*L# volume?