Rates of change (1 Viewer)

[fashionista]

hiii!! im having a bit of trouble with this rates of change question and i was really hoping someone could help me *big puss-in-boots eyes*

here goeeees:
When a gas expands without change of temp, the pressure p and volume v are given by the relation
pv^(1.4)=k , a constant.
At a certain instant, the pressure is 25 gm/cm^2 and the volume is 32 cm^3. If the volume is increasing at the rate of 5 cm^3/s, at what rate is the pressure changing at this instant??

i keep getting like 5600 gm/s or whatever the units are. golly gosh i be needing the helpings
luv me!!!

 

[CM_Tutor]

PV^1.4 = k

At P = 25 gcm^-2, V = 32 cm³: k = 25 * 32^1.4 = 3200

So, P = 3200 / V^1.4 = 3200V^-1.4
dP/dV = 3200 * -1.4V^-2.4 = -4480 / V^2.4

Now, we know that dV/dt = +5 cm³s^-1

Applying the Chain Rule, dP/dt = (dP/dV) * (dV/dt)
= (-4480 / V^2.4) * 5
= -22400 / V^2.4

So, at V = 32 cm³, dP/dt = -22400 / 32^2.4 = -22400 /4096 = -175 / 32 gcm^-2s^-1

In other words, the pressure is decreasing at a rate of 175 / 32 gcm^-2s^-1 at this instant.

 

[fashionista]

thank you!! i was going to post to say nevermind because i figrued out what i did wrong...when i was differentiating i added a power instead of subtracting silly silly me...but thankyou for helping!!!

 

[kpq_sniper017]

q10 from fitpatrick ex. 25(a)?
only from memory....but i remember my teacher saying it was an interesting question....didn't seem too interesting though