Question

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 0.14 atm/min and V = 13 L and is decreasing at a rate of 0.16 L/min. Find the rate of change of T with respect to time at that instant if n = 10 mol. (Round your answer to four decimal places.)dT/dt = ___ K/min

Answer 1

Answer:

$\frac{dT}{dt}=3.78^{\circ}K/min$

Step-by-step explanation:

We have to calculate the time derivative of T=PV/nR with P and V variable and n and R constants. This is:

$\frac{dT}{dt} =\frac{d\frac{PV}{nR}}{dt}=\frac{1}{nR}\frac{d(PV)}{dt}$

What we have to do is the derivative of a product:

$\frac{d(PV)}{dt}=P\frac{dV}{dt}+V\frac{dP}{dt}$

Substituting, we have:

$\frac{dT}{dt} =\frac{P\frac{dV}{dt}+V\frac{dP}{dt}}{nR}$

where all these values are given since the time derivatives of P and V are their variation rate, using minutes.

We then substitute everything, noticing that already everything is in the same system of units so they cancel out:

$\frac{dT}{dt}=\frac{P\frac{dV}{dt}+V\frac{dP}{dt}}{nR}=\frac{(8atm)(0.16L/min)+(13L)(0.14atm/min)}{(10mol)(0.0821Latm/mol^{\circ}K)}$

And then just calculate:

$\frac{dT}{dt}=3.78^{\circ}K/min$