An ideal gas undergoes an isothermal process starting with a pressure of 2 *10⁵ Pa and a volume of 6 cm³.Which of the following might be the pressure and volume of the final state? A)1 * 10⁵ Pa and 10cm³ B)3 * 10⁵ Pa and 6 cm³ C)4 * 10⁵ Pa and 4 cm³ D)6 * 10⁵ Pa and 2 cm³ E)8 * 10⁵ Pa and 2 cm³

Question

Quizplus · Accepted Answer

In an isothermal process, the temperature remains constant. Therefore, we can use the ideal gas law to find the final pressure and volume:

P1V1 = P2V2

where P1 = 2 x 10^5 Pa, V1 = 6 cm^3, and P2 and V2 are the final pressure and volume, respectively. Since the process is isothermal, we can also use the equation PV = nRT, where n is the number of moles of gas, R is the gas constant, and T is the temperature.

Setting the initial and final states equal, we have:

P1V1 = nRT

and

P2V2 = nRT

Dividing the second equation by the first, we get:

P2/P1 = V1/V2

Substituting the values we have:

P2/2 x 10^5 Pa = 6 cm^3/V2

P2V2 = 12 x 10^5 cm^3 Pa

Now, we can use the ideal gas law to find the final volume:

PV = nRT

Substituting P2 = 6 x 10^5 Pa, V2 = ?, and the other values:

6 x 10^5 Pa x V2 = n x R x T

Since the process is isothermal, T remains constant. Therefore, we can write:

6 x 10^5 Pa x V2 = n x R x T1

where T1 is the initial temperature.

We can simplify this equation using the fact that for an ideal gas, PV = nRT:

6 x 10^5 Pa x V2 = P1V1

Substituting the values we have:

6 x 10^5 Pa x V2 = 2 x 10^5 Pa x 6 cm^3

V2 = 2 cm^3

Therefore, the final state is:

Pressure = 6 x 10^5 Pa
Volume = 2 cm^3

The answer is choice D.

An Ideal Gas Undergoes an Isothermal Process Starting with a Pressure