The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at a certain instant, P = 8.0 atm and is increasing at a rate of 0.13 atm/min and V = 14 L and is decreasing at a rate of 0.17 L/min. Find the rate of change of T with respect to time at that instant if n = 10 moles.

PV = nRT

V dP/dt + P dV/dt = nR dT/dt

Now just plug in the data:

(14)(0.13) + (8.0)(-0.17) = (10)(0.0821) dT/dt
dT/dt = 0.56 °K/s

Well, it seems like we have a somewhat complex problem here, but don't worry, I'm here to make it less intimidating! Let's break it down step by step.

First, we need to find the rate of change of T with respect to time, so we'll be looking for dT/dt. To find this, we can use the given values and the chain rule of differentiation.

We know that PV = nRT, so let's differentiate this equation with respect to time t:

P dV/dt + V dP/dt = nR dT/dt

Now, we can plug in the given values and solve for dT/dt:

8.0 atm * (-0.17 L/min) + 14 L * (0.13 atm/min) = 10 mol * 0.0821 L atm / (mol K) * dT/dt

-1.36 L*atm/min + 1.82 L*atm/min = 0.821 L*atm/min*K * dT/dt

0.46 L*atm/min = 0.821 L*atm/min*K * dT/dt

Now, let's solve for dT/dt:

dT/dt = (0.46 L*atm/min) / (0.821 L*atm/min*K)

After doing the math, we find that:

dT/dt = 0.5611 K/min

So, the rate of change of temperature with respect to time at that instant is approximately 0.5611 K/min.

I hope this helps, and remember, always keep your sense of humor when dealing with gas laws!

To find the rate of change of temperature with respect to time, we need to differentiate the equation PV = nRT with respect to time, and solve for dT/dt.

The given information:
P = 8.0 atm (dP/dt = 0.13 atm/min, the rate of change of pressure with respect to time)
V = 14 L (dV/dt = -0.17 L/min, the rate of change of volume with respect to time)
n = 10 moles
R = 0.0821 L·atm/(mol·K)

Differentiating the equation PV = nRT with respect to time gives:
d(PV)/dt = d(nRT)/dt
P(dV/dt) + V(dP/dt) = nR(dT/dt)

Substituting the given values:
8.0 atm(-0.17 L/min) + 14 L(0.13 atm/min) = 10 mol(0.0821 L·atm/(mol·K))(dT/dt)

Simplifying the equation gives:
-1.36 atm·L/min + 1.82 atm·L/min = 0.821 L·atm/(K·min)(dT/dt)
0.46 atm·L/min = 0.821 L·atm/(K·min)(dT/dt)

Finally, rearranging the equation to solve for dT/dt gives:
dT/dt = (0.46 atm·L/min) / (0.821 L·atm/(K·min))
dT/dt = 0.46 * (K/min)
dT/dt = 0.46 K/min

Therefore, the rate of change of temperature with respect to time at that instant is 0.46 K/min.

To find the rate of change of temperature (dT/dt) with respect to time at a given instant, we need to use the concept of partial derivatives.

Given the gas law equation PV = nRT, we can rewrite it as:

P = (nR/V)T

Now, let's differentiate both sides of the equation with respect to time (t):

(dP/dt) = (d(nR/V)/dt) * T + (nR/V) * (dT/dt)

Since we are given the following values and their rates of change:

P = 8.0 atm, (dP/dt) = 0.13 atm/min
V = 14 L, (dV/dt) = -0.17 L/min
n = 10 moles

We can substitute these values into the above equation to find the rate of change of T with respect to time.

Substituting the given values:

0.13 = (d(nR/V)/dt) * T + (10 * 0.0821 / 14 ) * (dT/dt)

Simplifying:

0.13 = (d(nR/V)/dt) * T + 0.0472 * (dT/dt)

Now, we need to find the value of (d(nR/V)/dt). Since n and R are constant, we only need to focus on the rate of change of V.

Given:
(dV/dt) = -0.17 L/min

Therefore:

(d(nR/V)/dt) = (-nR/V^2) * (dV/dt)
= (-10 * 0.0821 / 14^2) * (-0.17)

Calculate the above expression:

(d(nR/V)/dt) = 0.0033 mol·atm·s/(L·K)

Substituting this value back into the equation:

0.13 = 0.0033 * T + 0.0472 * (dT/dt)

Now, rearrange the equation to solve for dT/dt:

0.0472 * (dT/dt) = 0.13 - (0.0033 * T)
(dT/dt) = (0.13 - (0.0033 * T))/0.0472

And that's the expression for the rate of change of temperature with respect to time at that instant.