Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio Vₐ/V_d and the ratio V_b/V_c is:

Solution

Physics sir · 4 steps

Use TV^(γ-1) = constant on each adiabat; both paths drop between the same two temperatures.

1. Read the graph. Points a and b lie on the upper isotherm at temperature T₁. Points d and c lie on the lower isotherm at T₂. One adiabat joins a to d, and the other joins b to c.
2. Adiabat a to d. For an adiabatic process TV^(γ-1) stays constant, so T₁ Vₐ^(γ-1) = T₂ V_d^(γ-1). This gives (Vₐ/V_d)^(γ-1) = T₂/T₁.
3. Adiabat b to c. In the same way, T₁ V_b^(γ-1) = T₂ V_c^(γ-1), which gives (V_b/V_c)^(γ-1) = T₂/T₁.

   (Vₐ/V_d)^(γ-1) = T₂/T₁ = (V_b/V_c)^(γ-1)

4. So. It is the same gas, so γ is the same on both paths. Taking the (γ-1)th root of both sides gives Vₐ/V_d = V_b/V_c, which is option (c).

The idea to keep. Any adiabat between the same two isotherms changes the volume by the same factor, (T₂/T₁)^(1/(γ-1)). This is the same reason the Carnot cycle gives V_b/Vₐ = V_c/V_d.

Answer from the NTA final answer key. Solution by PrepWiser.