#### To determine

**To find:** The rate of change of the volume of the gas when gas expands adiabatically (without gaining or losing heat).

#### Answer

The rate of change of the volume of the gas adiabatically is dVdt=2507≈36 cm3/min.

#### Explanation

**Given:**

The pressure P and volume V of the gas are related by the equation PV1.4=C, where C be any constant.

The rate at which the pressure of the gas decreases −10 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 PV1.4=C , where C be any constant.

Consider the pressure P of the gas decreases with the time t.

Therefore, the volume V of the gas become increases with the time t .

Since dPdt=−10 kPa/min

Find dVdt when P=80 kPa and V=400 cm3.

Differentiate PV1.4=C with respect to the time t.

ddt[PV1.4]=ddt[C]V1.4dPdt+Pddt[V1.4]=0 [Qddx(fg)=fdgdx+gdfdx]V1.4dPdt+P(1.4)V(1.4−1)dVdt =0 [Qdydx=dydu⋅dudx]V1.4dPdt+1.4PV0.4dVdt=0

On further simplification the rate of change of the volume of the gas adiabatically is as follows.

1.4PV0.4dVdt=−V1.4dPdtdVdt=−V1.41.4PV0.4⋅dPdtdVdt=−V(1.4−0.4)1.4P⋅dPdtdVdt=−V1.4P⋅dPdt

Substitutes P=80 ,V=400 and dPdt=−10 in dVdt .

dVdt=−4001.4×80×(−10)=501.4=50014=2507≈36 cm3/min

Therefore, the rate of change of the volume of the gas adiabatically is dVdt=2507≈36 cm3/min .

That is the volume of the gas increases at a rate of 36cm3/min.