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 0.13 atm/min and V = 13 L and is decreasing at a rate of 0.17 L/min. Find the rate of change of T with respect to time (in K/min) at that instant if n = 10 mol.

Answer 1

The rate of change of T with respect to time is 0.40 K/min

Step-by-step explanation:

The gas law equation is:

`PV = nRT`

We can find the rate of change of T with respect to time by solving the above equation for T and derivating with respect to time:

`(dT)/(dt) = (d)/(dt)((PV)/(nR))`

`(dT)/(dt) = (1)/(nR)(V(dP)/(dt) + P(dV)/(dt))`

Where:

n: is the number of moles = 10 mol

R: is the gas constant = 0.0821

V: is the volume = 13 L

P: is the pressure = 8.0 atm

dP/dt: is the variation of the pressure with respect to time = 0.13 atm/min

dV/dt: is the variation of the volume with respect to time = -0.17 L/min

Hence, the rate of change of T is:

`(dT)/(dt) = (1)/(10*0.0821)(13*0.13 - 8.0*0.17) = 0.40 K/min`

Therefore, the rate of change of T with respect to time is 0.40 K/min

I hope it helps you!