Question

2.5 moles of an ideal gas are expanded isothermally from 12 dm3 to 25 dm3 against a pressure of 3.0 bar. Calculate the work obtained.​

Answer 1

Answer:

n = 2.5 mol;

 V1 = 12 dm3;

 V2 = 25 dm3

 P = 3.0 bar;

 W = ?

 W = -Pext × (V2 – V1)

= – 3 × (25 – 12)

= -39 dm3 bar

∵ V1 dm3 = 100 J

∴ W = -39 × 100  = -3900 J

= -3.9kJ

Explanation:

Answer 2

Given: 2.5 moles of an ideal gas are expanded isothermally from 12 dm³ to 25 dm³ against a pressure of 3.0 bar.

To find: The work obtained.

Solution:

• With respect to the pressure exerted and change in volume of the ideal gas, work is given by the formula.

        $W = - P ( \delta V)$

• Here, W is the work done, P is the pressure and δV is the change in pressure of the gas.
• On replacing the terms with the values given in the question, the obtained equation is,

        $W = - (3) (25 - 12)$

             $= - 39 dm^{3} bar$

• On conversion into joule, the work is

        $-39 * 100 = 3900 J$

Therefore, the work obtained is 3900 J.