The compressibility k of a substance is defined as the fractional change in volu
ID: 1536548 • Letter: T
Question
The compressibility k of a substance is defined as the fractional change in volume of that substance for a given change in pressure: k = -1/V dV/dP (a) Explain why the negative sign in this expression ensures k is always positive. (b) Show that if an ideal gas is compressed isothermally, its compressibility is given by k_1 = 1/P. (Do this on paper. Your instructor may ask you to turn in this work.) (c) Show that if an ideal gas is compressed adiabatically, its compressibility is given by k_2 = 1/(y^p). (Do this on paper. Your instructor may ask you to turn in this work.) (d) Determine the value for k_1 for a monatomic ideal gas at a pressure of 2.70 atm. atm^-1 (e) Determine the value for k_2 for a monatomic ideal gas at a pressure of 2.70 atm. atm^-1Explanation / Answer
k = - dV/VdP [ where k is compressibility, V is volume, dV is fractional change in volume and dP os fractional change in pressure]
a. Now, if pressure is increased on a body, the compressive force would decrease its volume, and if the pressure is decreased on the body, the compressive -ve force ( outward pressure) would increase the volume of the bocy, henc e the sign of the numerator and the denominator on the right side of the equation are always different, hence a -ve will always be there, so an additional -ve sign ensures that k is always + ve
b. For isothermal compression
PV = constant
VdP + PdV = 0 ( differentiating both sides)
VdP = -PdV
1/P = -dV/VdP = k
k = 1/P ( where P is pressure, V is volume)
c. for adiabatic compression
PV^(gamma) = constant ( where gamma = heat capacity ratio of the gas)
so, dP*V^(gamma) + gamma*P*V^(gamma-1)dV = 0
dP = -PdV/V
k = -dV/VdP = 1/gamma*P
d. For P = 2.7 atm
k = 1/P = 0.3703 atm-1
e. for a monoatomic gas, gamma = 1.666
k = 1/gamma*P = 0.2223 atm-1
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