A gas undergoes two processes. In the first, the volume remains constant at 0.250 m^3 and the pressure increases from 2×10^5 Pa to 6×10^5 Pa. The second process is a compression to a volume of 8.00×10^(−2) m^3 at a constant pressure of 6×10^5 Pa.
Find the total work done by the gas during both processes in joules.
2006-06-26 22:25:33 · 7 answers · asked by Sagely 4 in Science & Mathematics ➔ Physics
W = PV = RTn, where P = pressure, V = volume, R = gas constant = 8.31 J/mol.K, T = temperature, n = moles of gas involved.
1) Change in P = 4 x 10^5 Pa, then:
Change in W = 4 x 10^5 Pa x 0.250 m^3 = 10^5 Pa.m^3 = 1 x 10^5 Joules.
2) Change in V = 17.00 x 10^(-2), then:
Change in W = 6 x 10^5 Pa x 17.00 x 10^(-2) m^3 = 102.0 x 10^3 Pa.m^3 = 1 x 10^5 Joules (note: there is only one significant number in result)
2006-06-26 22:42:51 · answer #1 · answered by tlakkamond 4 · 1⤊ 1⤋
Are you sure you're on the right website? Try a .edu site for questions like that.
BTW, Thanx for the 2 points. ; )>
2006-06-26 22:30:06 · answer #2 · answered by roxburger 3 · 0⤊ 1⤋
u should mention the kind of gas it is
that is if it is mono atomic or diatomic or polyatomic
that will give us the value of gamma. also u have not mentioned the temp of the gas
2006-06-26 22:55:28 · answer #3 · answered by klk 2 · 0⤊ 0⤋
Hey you think we are gonna do your homework???? too smart..
Do it yourself!
2006-06-26 22:33:18 · answer #4 · answered by gypsy gal 2 · 0⤊ 0⤋
0 and 1.452x10^5J
2006-06-26 22:32:19 · answer #5 · answered by cool 1 · 0⤊ 0⤋
42x25(pa)
2006-06-26 22:28:43 · answer #6 · answered by cocomademoiselle 5 · 0⤊ 0⤋
Jesus!!
2006-06-26 22:27:50 · answer #7 · answered by honky550 3 · 0⤊ 0⤋