Question

A gas cylinder holds 0.10 mol of O2 at 150$°$C and a pressure

of 3.0 atm. The gas expands adiabatically until the volume is doubled.

What are the final (a) pressure and (b) temperature?

Short answer

The final (a) pressure is 1.1368 atm and (b) temperature is 321 K.

Step-by-step solution

Preparing the bases of the formulas

(a) This is an adiabic process. So, it can be wriiten that

$$\begin{gathered} p_1{V_1}^{\gamma }=p_2{V_2}^{\gamma } \\ \frac{p_2}{p_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma } \\ {p}_2=p_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }\cdots \cdots \cdots \cdots \cdots \cdots \left(1\right) \end{gathered}$$

(b) The number of moles remains constant. From the ideal gas law,

$$\begin{gathered} \frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2} \\ {T}_2=\frac{p_2}{p_1}\frac{V_2}{V_1}{T}_1 \end{gathered}$$

and from part (a),

$\frac{p_2}{p_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }$

Substitutiong this equation, it can be continue finding,

$T_2={\left(\frac{V_1}{V_2}\right)}^{\gamma }\frac{V_2}{V_1}{T}_1={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}{T}_1\cdots \cdots \cdots \cdots \left(2\right)$

Putting the values

(a)$O_2$is a diatomic gas.

So, $\gamma =1.40$, Substituting this and the volume ratio, it can find from the equation (1) ,

$p_2=3·0.5^{1.40}=1.1368atm$

(b) By putting the numerical values on the equation (2),

$T_2=0.5^{1.40-1}·423=321K$