What is the rate of change of temperature with time in the following case:The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters)...

What is the rate of change of temperature with time in the following case:

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 = 12 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 mol. (Round your answer to four decimal places.)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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.

PV = nRT

Take the derivative of both the sides with respect to t

`V*(dP)/dt + P*(dV)/dt = n*R*(dT)/dt`

=> `(dT)/(dt) = (1/(n*R))*(V*(dP)/dt + P*(dV)/dt)`

At a certain instant of time, P = 8.0 atm and is increasing at a rate of 0.13 atm/min and V = 12 L and is decreasing at a rate of 0.17 L/min.

Substituting the values given

`(dT)/(dt) = (1/(10*0.0821))*(12*0.13 + 8*(-0.17))`

=> `(dT)/(dt) = (0.2/(10*0.0821))`

=> `(dT)/(dt) = 0.2436`

The rate of change of temperature with time is 0.2436 K/min

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