Question

The volume of a gas is halved during an adiabatic compression

that increases the pressure by a factor of 2.5.

a. What is the specific heat ratio$\gamma$?

b. By what factor does the temperature increase?

Short answer

a. The specific heat ratio is 1.32.

b. The temperature increase ration is 1.25.

Step-by-step solution

Formation of the base of the formulas

(a )The quantity remains same for an adiabatic process which means (considering two states as 1 and 2),

$$\begin{gathered} p_1{V_1}^{\gamma }=p_2{V_2}^{\gamma } \\ {\left(\frac{V_1}{V_2}\right)}^{\gamma }=\frac{p_2}{p_1} \\ \gamma ={\log }_{\left(\frac{V_1}{V_2}\right)}\left(\frac{p_2}{p_1}\right)\cdots \cdots \cdots \cdots \cdots \cdots \left(1\right) \end{gathered}$$

(b) As of the ideal gas law,

$$\begin{gathered} pV=nRT \\ \frac{pV}{T}=const. \end{gathered}$$

When, it is dealt with same amount of gas. This can written as,

role="math" localid="1649076704206" $\begin{array}{rcl}\frac{p_1{V}_1}{T_1}& =& \frac{p_2{V}_2}{T_2}\\ T_2& =& \frac{p_2{V}_2}{p_1{V}_1}{T}_1\dots \dots \dots \dots \left(2\right)\end{array}$

Solve of the problems

(a) $\gamma$is given parametrically, so, to find the substitute, as of the equation (1),

$\gamma ={\log }_{2}2.5=1.32$

(b) Let's the needed ratio be X.

So,role="math" localid="1649077113409" $X=\begin{array}{rcl}& & \frac{p_2{V}_2}{p_1{V}_1}\end{array}=\frac{2.5·\begin{array}{rcl}& & p\end{array}\begin{array}{rcl}& & \end{array}}{}$₁·0.5·V₁p1V1=1.25