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?

2 Answers
Oct 1, 2017

1430.678 Pa

Explanation:

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

Oct 1, 2017

Approx. 14*atm.....

Explanation:

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....