The pressure P and volume V of a gas are related as PV3/2 = K, where K is a constant. The percentage increase in the pressure for a diminution of 0.5 % of the volume is

Question

The pressure P and volume V of a gas are related as PV3/2 = K, where K is a constant. The percentage increase in the pressure for a diminution of 0.5 % of the volume is (A) 0.25 % (B) 0.75 % (C) 1.50 % (D) 0.50 %

Tardigrade Admin · Accepted Answer

As P propto V3/2 ⇒ (Δ P/P)=-(3/2) (Δ V/V). Therefore, percentage increase in pressure =(3/2)×0.5=0.75 %

Q. The pressure $P$ and volume $V$ of a gas are related as $P V^{3/2} = K$ , where $K$ is a constant. The percentage increase in the pressure for a diminution of $0.5%$ of the volume is

$0.25%$

$0.75%$

$1.50%$

$0.50%$

As $P \propto V^{3/2} \Rightarrow \frac{Δ P}{P} = - \frac{3}{2} \frac{Δ V}{V}$ . Therefore, percentage increase in pressure $= \frac{3}{2} \times 0.5 = 0.75%$

Q. The pressure P and volume V of a gas are related as PV3/2=K, where K is a constant. The percentage increase in the pressure for a diminution of 0.5% of the volume is

0.25%

0.75%

1.50%

0.50%

As P∝V3/2⇒PΔP​=−23​VΔV​. Therefore, percentage increase in pressure =23​×0.5=0.75%