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ayahouyee

New member
Nov 4, 2013
12
The ideal gas law states that PV = nRT where P is the pressure in atmospheres, V
is the volume in litres, n is the number of moles, R = 8.314 Latm/Kmol is the gas
constant, and T is the temperature in Kelvins. Suppose that at a specific instance that
two moles of gas is under 5 atmospheres of pressure where the pressure is decreasing at
0.3 atm/min. Also at this moment the volume is 15 L and is increasing at 0.6 L/min.
What is the rate of change of the temperature with respect to time?

Thanks in advance!
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
Look, you need to make an attempt at the solution or at least explain your difficulty.

Express T through P(t) and V(t). Then find the derivative dT/dt of T w.r.t. time treating P and V as functions of t.
 

ayahouyee

New member
Nov 4, 2013
12
Okay, so first i have to differentiate both sides with respect to time:

P(dV/dt) + V(dP/dt) = R(n(dT/dt) + T(dn/dt))

We're told that dP/dt = -0.3 atm/min, n = 2, P = 5 atm, V = 15 L, dV/dt = 0.6 L/min:

Plugging all this in:

(5 atm)(0.6 L/min) + (15 L)(-0.3 atm/min) = (8.314 L*atm/K*mol)*(2)(dT/dt)

This is where i get stuck! we are not allowed to use calculators so i dont know how i am supposed to find my final answer :(
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would do as suggested by Evgeny.Makarov and write:

\(\displaystyle T=\frac{1}{nR}PV\)

Now, the constant:

\(\displaystyle \frac{1}{nR}\)

will have units of:

\(\displaystyle \frac{1}{\text{mol}\cdot\frac{\text{J}}{\text{mol K}}}=\frac{\text{K}}{\text{J}}=\frac{\text{temperature}}{\text{energy}}\)

The two factors:

\(\displaystyle PV\)

will have units of:

\(\displaystyle \frac{\text{force}}{\text{length}^2}\cdot\text{length}^3=\text{force}\cdot\text{length}=\text{work}=\text{energy}\)

Thus, the equation is dimensionally consistent. The constant factor is:

\(\displaystyle \frac{1}{2\cdot8.314}=\frac{1}{16.628}=\frac{1000}{16628}=\frac{250}{4157}\)

So, we may now write:

\(\displaystyle T=\frac{250}{4157}PV\)

Now try differentiating with respect to time $t$.
 

ayahouyee

New member
Nov 4, 2013
12
DT/dt = 250/4157P(DP/dt) + 250/4157(DV/dt)

is that right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
DT/dt = 250/4157P(DP/dt) + 250/4157(DV/dt)

is that right?
No, you want to apply the product rule on the right side. Recall:

\(\displaystyle \frac{d}{dx}\left(f(x)g(x) \right)=f(x)\frac{d}{dx}\left(g(x) \right)+\frac{d}{dx}\left(f(x) \right)g(x)\)