/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 The molar specific heats of an i... [FREE SOLUTION] | 91Ó°ÊÓ

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The molar specific heats of an ideal gas at constant pressure and volume are denoted by \(C_{p}\) and \(C_{v}\), respectively. Further, \(\frac{C_{p}}{C_{v}}=\gamma\) and \(R\) is the gas constant for 1 gm mole of a gas. Then \(C_{v}\) is equal to (A) \(R\) (B) \(\gamma R\) (C) \(\frac{R}{\gamma-1}\) (D) \(\frac{\gamma R}{\gamma-1}\)

Short Answer

Expert verified
The short answer is \(C_v = \frac{R}{\gamma - 1}\), which corresponds to option (C).

Step by step solution

01

Write the equation for the gas constant R

We start by noting that the difference in molar specific heat at constant pressure and constant volume is equal to the gas constant. This can be written as: \(C_p - C_v = R\)
02

Use the given relationship between \(C_p\) and \(C_v\)

We are given that the ratio of \(C_p\) to \(C_v\) is equal to \(\gamma\). This can be written as: \(C_p = \gamma C_v\)
03

Substitute the relationship found in Step 2 into the equation in Step 1

Substituting the expression for \(C_p\) in terms of \(C_v\) found in the previous step into the equation in Step 1, we get: \(\gamma C_v - C_v = R\)
04

Solve for \(C_v\) in terms of R and \(\gamma\)

We can arrange the equation found in Step 3 to isolate \(C_v\): \(C_v (\gamma - 1) = R\) Now, we can divide both sides of the equation by \((\gamma - 1)\) to find the value of \(C_v\): \(C_v = \frac{R}{\gamma - 1}\) Therefore, the correct answer is \(C_v = \frac{R}{\gamma - 1}\), which corresponds to option (C).

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