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The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by the equatic PV = 8.317. Find the rate at which the pressure is changing when the temperature is 400 K and increasing at a rate of 0.1 K/s and the volume is 100 L and increasing at a rate of 0.2 L/s. Solution If t represents the time elapsed in seconds, then at the given instant, we have T = 400, P = 8.31 the chain rule gives the following. (Round your final answer to five decimal places.) dp aP dT aP dv dt + at dt av dt 11 8.31 dT V dt 8.31 100 dv dt 8.31(400) 100² dT = 0.1, V = 100, dv dt dt The pressure is decreasing at a rate (in kPA/s), rounded to three decimal places, of about KPA/s. = 0.2. Since

The pressure $P$ (in kilopascals), volume $V$ (in liters), and temperature $T$ (in kelvins) of a mole of an ideal gas are related by the equat $PV=8.31T$. Find the rate at which the pressure is changing when the temperature is $400K$ and increasing at a rate of $0.1K/s$ and th volume is $100L$ and increasing at a rate of $0.2L/s$. Solution If $t$ represents the time elapsed in seconds, then at the given instant, we have $T=400,dtdT =0.1,V=100,dtdV =0.2.5$ ince $P=8.31VT $ the chain rule gives the following. (Round your final answer to five dedimal places.) $dtdP =∂T∂P dtdT +∂V∂P ∂tdV =V8.31 dtdT −(=1008.31 ()dtdV = $ The pressure is decreasing at a rate (in kPA/s), rounded to three decimal places, of about $kPA/s$.

The pressure , volume , and temperature are given.

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