Question

The pressure p and volume V of a gas are related as pv^3/2=k .where K is a constant. the percentage increase in the pressure for the diminution of 0.5% of the volume is

Answer 1

The pressure P and volume V of a gas are related as $PV^{3/2}=k$. where k is a constant.

it is given that diminution of volume = 0.5% i.e., (∆v/v) × 100 = -0.5% [ here negative sign Indicates that volume is decreasing]

or you can say that percentage error of v = -0.5 %

from above relation,

% error formula will be ,

∆P/P = -(3/2) × ∆v/v

or, ∆P/P × 100 = -(3/2) × ∆v/v × 100

or, % error of p = -1.5 × % error of v

= -1.5 × (- 0.5 )

= + 0.75 % [ positive sign indicates pressure is increasing]

hence, percentage increase in the pressure = 0.75%

Answer 2

The ideal gas law can be written in terms of the number of molecules of gas: PV = NkT, where P is pressure, V is volume, T is temperature, N is number of molecules, and k is the Boltzmann constant k = 1.38 × 10–23 J/K. A mole is the number of atoms in a 12-g sample of carbon-12.