Question

A reversible adiabatic path on a P-V diagram for an ideal gas passes through state A where $ P = 0.7 \times {10^5}N{m^{ - 2}} $ and $ V = 0.0049{m^3} $ . The ratio of specific heat of gas is $ 1.4 $ . The slope of path at A is:
(A) $ 2.0 \times {10^7}N{m^{ - 5}} $
(B) $ 1.0 \times {10^7}N{m^{ - 5}} $
(C) $ - 2.0 \times {10^7}N{m^{ - 5}} $
(D) $ - 1.0 \times {10^7}N{m^{ - 5}} $

Answer 1

Hint: From the equation of an ideal gas in an adiabatic path, we can find the slope of the path by finding the ratio of change in pressure to the change in volume. In that after substituting the values given in the question, we get the slope.

Formula used: In the solution we will be using the following formula,
$\Rightarrow P{V^\gamma } = {\text{constant}} $
where $ P $ is the pressure, $ V $ is the volume and $ \gamma $ is the specific heat.

Complete step by step solution:
In the question we are given that an ideal gas follows an adiabatic path. So the ideal gas equation in a adiabatic path is given as,
$\Rightarrow P{V^\gamma } = {\text{constant}} $
So on differentiating on both the sides we get,
$\Rightarrow d\left( {P{V^\gamma }} \right) = 0 $ as the differentiation of a constant is zero.
So we get,
 $\Rightarrow {V^\gamma }dP + Pd\left( {{V^\gamma }} \right) = 0 $
On doing the differentiation of the volume we get,
 $\Rightarrow {V^\gamma }dP + P\gamma {V^{\gamma - 1}}dV = 0 $
Now taking one term to the RHS we get,
 $\Rightarrow {V^\gamma }dP = - P\gamma {V^{\gamma - 1}}dV $
Now, on taking $ dV $ to the LHS and the $ {V^\gamma } $ to the RHS we get,
 $\Rightarrow \dfrac{{dP}}{{dV}} = - \dfrac{{P\gamma {V^{\gamma - 1}}}}{{{V^\gamma }}} $
Hence we have,
 $\Rightarrow \dfrac{{dP}}{{dV}} = - P\gamma {V^{\gamma - 1 - \gamma }} $
So on simplifying we have,
 $\Rightarrow \dfrac{{dP}}{{dV}} = - \gamma \dfrac{P}{V} $
This is the slope of the path. Now in the question we are given $ P = 0.7 \times {10^5}N{m^{ - 2}} $ , $ V = 0.0049{m^3} $ and $ \gamma = 1.4 $
So substituting we get,
 $\Rightarrow \dfrac{{dP}}{{dV}} = - 1.4 \times \dfrac{{0.7 \times {{10}^5}}}{{0.0049}} $
On calculating we get, $ \dfrac{{dP}}{{dV}} = - 2 \times {10^7} $
Hence the slope is $ - 2.0 \times {10^7}N{m^{ - 5}} $
So the correct answer is option (C).

Note:
In thermodynamics an adiabatic process is that process which occurs without transferring the heat and mass between the system and the surroundings. The adiabatic process transfers energy to the surrounding in form of work.