Question

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.120 of air at a pressure of 3.50 .

The piston is slowly pulled out until the volume of the gas is increased to 0.430 . If the temperature remains constant, what is the final value of the pressure?

Answer 1

Concepts and reason

The concept of ideal gas equation is required to solve this problem.

Initially obtain expression for gas at two different states. Then rearrange equation for the pressure after the expansion of the gas. Finally substitute values to get pressure.

Fundamentals

The ideal gas equation is as follows:

$pV = nRT$ …… (1)

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

Consider initially pressure was ${p_1}$ and volume was ${V_1}$ , after expansion the pressure becomes ${p_2}$ and volume becomes ${V_2}$ . Hence, equation can be written for these two situations as follows:

$\begin{array}{l}\\{p_1}{V_1} = nRT\\\\{p_2}{V_2} = nRT\\\end{array}$

Left hand side of these two equations are same hence,

${p_1}{V_1} = {p_2}{V_2}$

Rearrange for ${p_2}$ .

${p_2} = \frac{{{p_1}{V_1}}}{{{V_2}}}$ …… (2)

Substitute 3.50 Pa for ${p_1}$ , 0.120 L for ${V_1}$ , and 0.430 L ${V_2}$ in equation (2).

$\begin{array}{c}\\{p_2} = \frac{{\left( {3.50{\rm{ Pa}}} \right)\left( {0.120{\rm{ L}}} \right)}}{{\left( {0.430{\rm{ L}}} \right)}}\\\\ = 0.98{\rm{ Pa}}\\\end{array}$

Ans:

The final pressure of the gas is 0.98 Pa.