A sample of gas (24.2 g)initially at 4.00 atm was compressed from 8.00 L to 2.00 L at constant temperature.After the compression,the gas pressure was ________ atm.
A)4.00
B)2.00
C)1.00
D)8.00
E)16.0

Question

Quizplus · Accepted Answer

Using the ideal gas law, we can solve for the final pressure:

PV = nRT

n = m/MW (where n is the number of moles, m is the mass, and MW is the molar mass)

MW = 28.01 g/mol (assumed to be air)

n = 24.2 g / 28.01 g/mol = 0.864 mol

R = 0.08206 L⋅atm/mol⋅K

T is constant, so we can write:

P1V1 = nRT   and   P2V2 = nRT

Combining these equations, we get:

P1V1/P2V2 = 1

P2 = P1V1/V2 = (4.00 atm)(8.00 L)/(2.00 L) = 16.0 atm

Therefore, the gas pressure after compression is 16.0 atm. The best choice is E.

A Sample of Gas (24