Question

Two cylinders each contain $0.10mol$of a diatomic gas at $300\mathrm{K}$and a pressure of $3.0\mathrm{atm}$. Cylinder A expands isothermally and cylinder B expands adiabatically until the pressure of each is $1.0\mathrm{atm}$.

a. What are the final temperature and volume of each?

b. Show both processes on a single $pV$diagram. Use an appropriate scale on both axes.

Short answer

a. The final temperature and volume of each cylinder is $300\mathrm{K}$and $2.469\mathrm{L}$respectively.

b. The$pV$diagram is as follows:

Step-by-step solution

Given Information (Part a)

Amount of diatomic gas in two cylinders $=0.10mol$

Temperature$=300K$

Pressure$=3.0atm$

Cylinder A expands isothermally till the pressure reach$1.0atm$

Cylinder B expands adiabatically till the pressure reach$1.0atm$

Explanation (Part a)

(a) Since cylinder A expands isothermally, by definition, its temperature after the expansion will be the same as before; that is$300K$.

Knowing that the expansion was isothermic, we have

$p_1{V}_1=p_2{V}_2\Rightarrow V_2=\frac{p_1}{p_2}{V}_1$

The initial volume $V_1$can be found from the Ideal Gas Law:

$p_1{V}_1=nR{T}_1\Rightarrow V_1=\frac{nR{T}_1}{p_1}=\frac{0.1·8.314·300}{3·1.01·{10}^5}=\underset{¯}{8.23·{10}^{-4}{\mathrm{m}}^3}$

Therefore, the final volume of cylinder A will be

$V_2=3·V_1=3·0.823=2.469\mathrm{L}$

For cylinder B we have an adiabatic process, which means that

$p_1{V}_1^{\gamma }=p_2{V}_2^{\gamma }$

We need to find an expression for $V_2$.

We have,

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

Finally,

$V_2=V_1\sqrt{\frac{p_1}{p_2}}$

For diatomic gases we have $\gamma =1.40$, and we know that the pressure ratio is $p_1/p_2=3$, as well as the fact that the initial volume was $0.823$Liters.

We therefore can calculate numerically,

$V_2=0.823·\sqrt[1.4]{3}=1.80\mathrm{L}$

Final Answer (Part a)

As for the temperature, from the Ideal Gas Law, we can find that,

$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}\Rightarrow T_2=\frac{p_2{V}_2}{p_1{V}_1}{T}_1=\frac{1·1.8}{3·0.823}·300=219\mathrm{K}$

Given Information (Part b)

Amount of diatomic gas in two cylinders$=0.10\mathrm{mol}$

Temperature$=300\mathrm{K}$

Pressure$=3.0\mathrm{atm}$

Cylinder A expands isothermally till the pressure reach$1.0\mathrm{atm}$

Cylinder B expands adiabatically till the pressure reach$1.0\mathrm{atm}$

Explanation (Part b)

An approximate graphical representation of the situation is given below:

Final Answer(Part b)

Therefore, the diagram is as follows: