Difficulty: Easy
Correct Answer: Both (b) and (c)
Explanation:
Introduction / Context:
For ideal gases with temperature-independent specific heats over a range, cp, cv, γ, and the gas constant R are linked by simple identities used throughout thermodynamics and gas dynamics.
Given Data / Assumptions:
Concept / Approach:
From the ideal-gas relations and definitions, two cornerstone identities follow: cp − cv = R and cp/cv = γ. Combining them yields convenient forms for cp and cv in terms of R and γ: cp = γR/(γ − 1) and cv = R/(γ − 1).
Step-by-Step Solution:
Start with cp − cv = R.Use γ = cp/cv ⇒ cp = γcv.Substitute into cp − cv = R → γcv − cv = R → cv(γ − 1) = R.Solve: cv = R/(γ − 1) and cp = γR/(γ − 1).
Verification / Alternative check:
Dimensional check: cp, cv, and R share units J/(kg·K) (for specific values); formulas are dimensionally consistent. Plugging back reproduces cp − cv = R and γ = cp/cv.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing molar vs. specific (per mass) quantities; the same relations hold for both as long as R is the corresponding gas constant (Rmolar or Rspecific).
Final Answer:
Both (b) and (c)
Discussion & Comments