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 = 9.0 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.)

Answer 1

Answer:

The rate of change of the temperature is 0.0365 Kelvin per minute.

Explanation:

Step 1: Given data

ideal gas law: P*V = n*R*T

with P= pressure of the gas ( in atm) = 9.0 atm

with V= volume of the gass (in L) =12L

with n = number of moles = 10 moles

R = gas constant = 0.0821 L*atm* K^−1*mo^−1

T = temperature = TO BE DETERMINED

The volume  decreases with a rate of 0.17L/min = dV/dT  = -0.17

The pressure increases at a rate of 0.13atm/min = dP/dT

Step 2: The ideal gas law

P * [dV/dT] + V * [dP/dT] = nR * dT/dt

9 atm * (-0.17L/min) + 12L * 0.13atm/min = 10 moles * 0.0821 L*atm* K^−1*mo^−1 *dT/dt

0.03 = 0.821 * dT/dt

dT/dt = 0.03/0.821

dT/dt = 0.0365

Since the gas constant is expressed in Kelvin and not in °C, this means that the rate of chagnge of the temperature is 0.0365 Kelvin per 1 minute.