Question

Slope of isotherm for a gas is 3×10^5.if the same gas is undegoing adiabatic change then adiabatic elasticity at that instant is​

Answer 1

Given:

The isothermal for gas is P= 3×10⁵

γ=1.4

To find:

The adiabatic elasticity of the gas

Solution:

The formula of the isothermal and adiabatic is

E=γP

Putting the above values

E=1.4×3×10⁵

E=1.4×10⁵

So the adiabatic elasticity is 1.4×10⁵

Answer 2

Answer:

5 x 10⁵ Nm⁻²

Explanation:

For an Isothermal Process

PV = constant

If you differentiate,

PdV + VdP = 0

∴  $\frac{dP}{dV }$ =   $-\frac{P}{V\\}$

For an Adiabatic Process

$PV^{5/3}$ = constant

∴ $\frac{dP}{dV} =$ $-\frac{5}{3} \frac{P}{V}$

Thus

Adiabatic slope or Adiabatic elasticity = γ x isotherm slope

= $\frac{5}{3}$ x 3x 10⁵ = 5 x 10⁵ N/m²