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Problem 37 Hard Difficulty

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 $ P = 0.0821 $ is the gas constant. Suppose that, at a certain instant, $ P = 8.0 $ atm and is increasing at a rate of $ 0.10 atm/min $ and $ V = 10 L $ and is decreasing at a rate of 0.15 L/min. Find the rate of change of $ T $ with respect to time at that instant if $ n = 10 mol.

Answer

$-0.2436 \mathrm{K} / \mathrm{min}$

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Video Transcript

here we have the classic equation. PV equals NRT the ideal gas law. And our goal is to find the rate of change of tea with respect to time. So we want to find duty. DT. So let's take our equation and isolate capital t So we have capital t equals P V divided by in our now in and are are constants and PNV are variables. So maybe we should write it as one over in our times. The product p Times V and when we differentiate will use the product rule. We have values for a constants in and are so we can substitute those into our equation. So T equals one over 10 times 0.8 to 1. We could definitely simplify that now or later. And then we have that multiplied by p times beat. So now we're ready to find the derivative. So again we're going to use the product rule. We keep the constant here. That's gonna be 1/0 0.821 and then we multiply by the first p times, the derivative of the second DVD T plus the second V times, the derivative of the first to dp DT And now we were given some numbers for a particular instance. In time, we have a pressure and we have ah rate of change of pressure. Ah, volume and a rate of change of volume. So we can substitute those numbers in. So we have 1/0 0.8 to 1 times eight time 0.15 plus 10 10 0.10 and then you compute all of that and you end up with approximately negative 0.244 And the units on temperature were Kelvin and the units on time. We're minutes. So negative 0.2 for four Calvin per minutes per minute.

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