Question

A gas is confined to a cylindrical volume with a movable piston on one end. Initiall, the volume of the container is 0.160 m^3, and the gas is at 315 K. The gas is heated, but the piston is moved such that the temperature of the gas remains constant.

If there are 1.60 moles of gas in the container, write an equation for the pressure of the gas as a function of the volume of the container. (Assume SI units, and do not include units in your answer. Use the variable, V, as necessary.)

Answer 1

Solution:

In the above given arrangement, the gas is heated but the piston is moved such that temperature of the gas remains constant. Thus it is an isothermal process. The diagram shows the graph of pressure of the gas versus the volume of the gas while the temperature is kept constant.

These curves are known as isotherms and they corresponds to a particular temperature. One the isotherm is drawn at temperature of 315K, it shows how pressure of the gas changes with the change in volume.

We have the ideal gas equation

pV = nRT                ---------------------- (1)

where p = pressure of the gas

V= volume of the gas

R = Gas constant = 8.31 J/mol.K

n = number of moles of the gas

T= temperature of the gas

Thus rearranging equation for pressure, we have

p = nRT/V            -----------------------(2)

Equation (2) represents the equation for pressure as function of gas. We note that number of moles (n), Gas constant (R) and Temperature of gas (T) are all constants. Thus they are a unchanging number.

Thus from equation (2)

p = (1.6*8.31*315)/V

p = 4188.24/V

This is an equation for the pressure of the gas as a function of the volume of the container, hence required answer.