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

Answer:

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

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:

$\frac{dT}{dt} = \frac{d}{dt}(\frac{PV}{nR})$

$\frac{dT}{dt} = \frac{1}{nR}(V\frac{dP}{dt} + P\frac{dV}{dt})$

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:

$\frac{dT}{dt} = \frac{1}{10*0.0821}(13*0.13 - 8.0*0.17) = 0.40 K/min$    

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

I hope it helps you!