Question:
Which one is the correct option for the two different thermodynamic processes?
Which one is the correct option for the two different thermodynamic processes?
Show Hint
Check each graph: isothermal means constant T (a vertical line on a plot with T on the x-axis; \(PV\) constant on P-V); adiabatic has \(PV^\gamma\), \(TV^{\gamma-1}\) and \(P^{1-\gamma}T^{\gamma}\) constant.
Updated On: Nov 4, 2026
- (a) only
- (c) and (d)
- (b) and (c)
- (c) and (a)
Show Solution
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The Correct Option is B
Solution and Explanation
(a) P-V graph: a vertical line (V constant) is an isochoric process, not adiabatic. So (a) is wrong.
(b) P-T graph: a horizontal line (P constant) is an isobaric process, not isothermal. So (b) is wrong.
(c) V-T graph: a vertical line (T constant) is isothermal; for an adiabatic process \(TV^{\gamma-1} = \mathrm{const}\), so V falls as T rises, a decreasing curve. So (c) is correct.
(d) P-T graph: a vertical line (T constant) is isothermal; for an adiabatic process \(P^{1-\gamma}T^{\gamma} = \mathrm{const}\), so \(P \propto T^{\gamma/(\gamma-1)}\), an increasing convex curve. So (d) is correct.
Correct graphs: (c) and (d).
Answer: (2) (c) and (d)
(b) P-T graph: a horizontal line (P constant) is an isobaric process, not isothermal. So (b) is wrong.
(c) V-T graph: a vertical line (T constant) is isothermal; for an adiabatic process \(TV^{\gamma-1} = \mathrm{const}\), so V falls as T rises, a decreasing curve. So (c) is correct.
(d) P-T graph: a vertical line (T constant) is isothermal; for an adiabatic process \(P^{1-\gamma}T^{\gamma} = \mathrm{const}\), so \(P \propto T^{\gamma/(\gamma-1)}\), an increasing convex curve. So (d) is correct.
Correct graphs: (c) and (d).
Answer: (2) (c) and (d)
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