#### To determine

**To find:** The rate of change of the volume of the gas at constant temperature.

#### Answer

The rate of change of the volume of the gas is dVdt=−80 cm3/min .

#### Explanation

**Given:**

The pressure P and volume V of the gas are related by the equation PV=C, where C be any constant.

The rate at which the pressure of the gas increases 20 kPa/min.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

(2) ddx(fg)=fdgdx+gdfdx .

**Calculation:**

Since the pressure P of the gas and volume V of the gas related by the equation PV=C , where C be any constant.

Since the pressure P of the gas increases with the time t.

Therefore the volume V of the gas deceases with the time t .

Since dPdt=20 kPa/min

Find dVdt when P=150 kPa and V=600 cm3.

Differentiate PV=C with respect to the time t.

ddt[PV]=ddt[C]VdPdt+PdVdt=0 [Qddx(fg)=fdgdx+gdfdx]PdVdt=−VdPdtdVdt=−VP⋅dPdt

Substitutes P=150 ,V=600 and dPdt=20 in dVdt .

dVdt=−600150×20=−4×20=−80 cm3/min

Therefore, the rate of change of the volume of the gas is dVdt=−80 cm3/min .

That is the volume of the gas decreases at a rate of 80cm3/min.