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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 = 9.0 atm and is increasing at a rate of 0.13 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.)

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Answer:

T is increasing at a rate of 0.3045 K/min.

Step-by-step explanation:

PV = nRT

nRT = PV

(d/dt)(nRT) = (d/dt) (PV)

Since nR are constants, we have

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

n = 10 moles

R = 0.0821 atm.L/mol.K

P = 9 atm

V = 13 L

(dP/dt) = 0.13 atm/min

(dV/dt) = -0.16 atm/min

(dT/dt) = ?

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

(10)(0.0821)(dT/dt) = (9)(-0.16) + (13)(0.13)

0.821 (dT/dt) = 0.25

(dT/dt) = (0.25/0.821)

(dT/dt) = 0.3045 K/min

Hope this helps!!

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