Ideal Gas Law: Pressure, Volume, & Temperature Changes

[ayahouyee]

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 L
atm/K
mol 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]

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]

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 don't know how i am supposed to find my final answer :(

 

[MarkFL]

I would do as suggested by Evgeny.Makarov and write:

$$T=\frac{1}{nR}PV$$

Now, the constant:

$$\frac{1}{nR}$$

will have units of:

$$\frac{1}{\text{mol}\cdot\frac{\text{J}}{\text{mol K}}}=\frac{\text{K}}{\text{J}}=\frac{\text{temperature}}{\text{energy}}$$

The two factors:

$$PV$$

will have units of:

$$\frac{\text{force}}{\text{length}^2}\cdot\text{length}^3=\text{force}\cdot\text{length}=\text{work}=\text{energy}$$

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

$$\frac{1}{2\cdot8.314}=\frac{1}{16.628}=\frac{1000}{16628}=\frac{250}{4157}$$

So, we may now write:

$$T=\frac{250}{4157}PV$$

Now try differentiating with respect to time $t$.

 

[ayahouyee]

DT/dt = 250/4157P(DP/dt) + 250/4157(DV/dt)

is that right?

 

[MarkFL]

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:

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