Question

A flask with a volume of 1.50 L , provided with a stopcock, contains ethane gas 1C2H62 at 300 K and atmospheric pressure$\mathbf{11}\mathbf{.}\mathbf{013}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathit{P}$ . The molar mass of ethane is$\mathbf{30}\mathbf{.}\mathbf{1}\mathbf{}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$ . The system is warmed to a temperature of , with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

Short answer

The final pressure of the ethane in the flask is$1.688\times {10}^{5}Pa$ and the remain ethane in flask is 167 g .

Step-by-step solution

Definition of ethane

The term ethane is defined as the carbon compound having two carbon elements and six hydrogen elements

Determine the final pressure of the ethane in the flask and ethane remain in the flask

Using the ideal gas law

$PV=nRT$

Here, p is the pressure, is the volume, V is the number of moles, R is the universal ideal gas constant, and T is the temperature.

Here, n , V , and R are constant so the relation between temperature and pressure for two states is,

$\frac{p_1}{p_2}=\frac{T_1}{T_2}$ ….. (1)

Consider the given data as below.

The pressure for the first state,$p_1=1.013\times {10}^{5}Pa$

The temperature of the first state,$T_1=330K$

The pressure of the second state is$p_2$ .

The temperature of the second state,$T_2=550K$

The volume,$V=1.5L=1.5\times {10}^{-3}{m}^3$

The molar mass,$M=30.19g/mol$

Substitute these values into equation (1), and you obtain

$$\begin{gathered} P_2=\frac{P_1{T}_2}{T_1} \\ {P}_2=\frac{1.013\times {10}^5\times 550}{330} \\ =1.688\times {10}^{5}Pa \end{gathered}$$

Hence, the final pressure of the ethane in the flask is$1.688\times {10}^{5}Pa$ .

Now,the number of moles can be written as,

$n=\frac{m}{M}$

Where m is the mass of gas given and M is the atomic weight.

Now using ideal gas equation

$$\begin{gathered} pV=\frac{m}{M}RT \\ m=\frac{p_{2}VM}{R{T}_2} \\ m=\frac{1.688\times {10}^5\times 1.5\times {10}^{-3}\times 30.19}{8.3145\times 550} \\ =1.67g \end{gathered}$$

Hence, the remain ethane in flask is 167 g .