Question

Slope of isotherm for a gas (having $\gamma =\dfrac{5}{3}$) is $3\times {{10}^{5}}N{{m}^{-2}}$. If the same gas is undergoing adiabatic change then adiabatic elasticity at that instant is –

& \text{A) }3\times {{10}^{5}}N{{m}^{-2}} \\\ & \text{B) 5}\times {{10}^{5}}N{{m}^{-2}} \\\ & \text{C) 6}\times {{10}^{5}}N{{m}^{-2}} \\\ & \text{D) 10}\times {{10}^{5}}N{{m}^{-2}} \\\ \end{aligned}$$

Answer 1

We need to find the relation between the isothermal and adiabatic process for the same gas at a constant temperature. We can find the adiabatic elasticity or the slope of the adiabatic process from the relation between the two processes.

Complete answer:
We are given the adiabatic index of a gas which undergoes isothermal and adiabatic change at some time instants. Adiabatic index is the ratio of specific heat capacities at constant and constant volume.
We know that in an isothermal the product of the pressure and volume will be a constant. The isothermal slope is the slope in the Pressure versus volume graph in an isothermal change of an ideal gas.
The isothermal condition is given by –
$PV=\text{constant}$
Now, let us partially differentiate the above relation to get the slope of the pressure versus volume graph for an isothermal change as –

& PdV+VdP=0 \\\ & \Rightarrow \dfrac{dP}{dV}=-\dfrac{P}{V} \\\ \end{aligned}$$ Now, let us consider the adiabatic change in the same gas at the same temperature. We know that, for adiabatic changes the pressure and volume are related by the adiabatic index, $$\gamma $$. The slope of the pressure versus volume graph of an adiabatic change is the adiabatic slope or the adiabatic elasticity. We can write the relation as – $$P{{V}^{\gamma }}=\text{constant}$$ Now, we can differentiate this relation to get the slope of the adiabatic change as – $$\begin{aligned} & P{{V}^{\gamma }}=\text{constant} \\\ & \Rightarrow \gamma P{{V}^{\gamma -1}}dV+{{V}^{\gamma }}dP=0 \\\ & \Rightarrow \gamma P{{V}^{\gamma -1}}dV=-{{V}^{\gamma }}dP \\\ & \Rightarrow \dfrac{dP}{dV}=-\dfrac{\gamma P}{V} \\\ \end{aligned}$$ Now, we can find the adiabatic elasticity by comparing the slopes of the isothermal and adiabatic conditions as – $$\begin{aligned} & {{\dfrac{dP}{dV}}_{isothermal}}=-\dfrac{P}{V}=3\times {{10}^{5}} \\\ & \text{and,} \\\ & \gamma =\dfrac{5}{3}, \\\ & \Rightarrow {{\dfrac{dP}{dV}}_{adiabatic}}=-\dfrac{\gamma P}{V} \\\ & \Rightarrow {{\dfrac{dP}{dV}}_{adiabatic}}=\dfrac{5}{3}\times 3\times {{10}^{5}} \\\ & \therefore {{\dfrac{dP}{dV}}_{adiabatic}}=5\times {{10}^{5}}N{{m}^{-2}} \\\ \end{aligned}$$ We get the adiabatic elasticity of the gas to be $$5\times {{10}^{5}}N{{m}^{-2}}$$. **The correct answer is option B.** **Note:** The adiabatic process involves an additional factor of the adiabatic index in the slope calculation of an ideal gas. The adiabatic index is the ratio of the specific heat capacity of the gas at constant pressure to that of at a constant volume for an ideal gas.