2006-12-23 00:59:12 · 9 answers · asked by Yorkie1961 2 in Science & Mathematics ➔ Mathematics
dP/dV=-nRT/V^2
2006-12-23 01:01:25 · answer #1 · answered by raj 7 · 3⤊ 1⤋
dp/dv is the rate of change of p with respect to v.
Then by simple differentiation treating nRT as constant and v as the variable with respect to which we are differentiating.
p= nRT/v
dp/dv = (nRT)d(V^-1)/dv = nRT.[-(v^-2)] = -nRT / V^2
Hope this helps!
2006-12-23 02:36:08 · answer #2 · answered by yasiru89 6 · 1⤊ 0⤋
P = nRT/V
now, dp/dv=0
bcoz,differentiating a const gives u zero.
2006-12-23 01:02:19 · answer #3 · answered by For peace 3 · 0⤊ 2⤋
Think of it this way:
P(V) = nRT* V^-1
So,
dP/dV= -1 * nRT * V^-2 = -nRT/(V^2)
Done.
2006-12-23 02:54:26 · answer #4 · answered by Jerry P 6 · 0⤊ 0⤋
(nrt= constant =c)
p= c/v
dp/dv= (c/v)'
dp/dv= [(c'*v)-(c*v')]/(v^2) and c'=0
dp/dv = -(c*v')/(v^2) = -(nrt*v')/(v*v)
edit : (v'=1)
-----> -(nrt*v')/(v*v) = -(nrt)/(v*v)
2006-12-23 01:20:56 · answer #5 · answered by fbalpfb 2 · 0⤊ 1⤋
dp/dv=-nrt/v*v
2006-12-23 01:07:07 · answer #6 · answered by JAMES 4 · 0⤊ 1⤋
dP/dV=-nRT/V^2
2006-12-23 06:18:00 · answer #7 · answered by mu_do_in 3 · 0⤊ 0⤋
-1/(v*v)
2006-12-23 01:07:14 · answer #8 · answered by pikus_zeech 2 · 0⤊ 1⤋
-nRT/v^2
2006-12-23 01:18:12 · answer #9 · answered by openpsychy 6 · 0⤊ 1⤋