Answer:
Under isothermal conditions, the bulk modulus of elasticity is \[{{K}_{iso}}=\frac{\Delta P}{\Delta {{V}_{1}}/V}=P\] ... (i) Under adiabatic conditions, the bulk modulus of elasticity is \[{{K}_{adia}}=\frac{\Delta P}{\Delta {{V}_{2}}/V}=\gamma P\] ?(ii) Dividing (ii) by (i), we get \[\frac{\Delta {{V}_{1}}}{\Delta {{V}_{2}}}=\gamma \] As \[\gamma >1,\]so \[\Delta {{V}_{1}}>\Delta {{V}_{2}}\]
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