Question

What will be the work done when one mole of a gas
expands isothermally from 15 L to 50 L against a
constant pressure of 1 atm at 25 °C?
(a) - 3542 cal
(b) - 843.3 cal
(c) - 718 cal
(d) - 60.23 cal​

Answer 1

ANSWER :

As we v know that, work done in isothermal expansion is :-

✒ W = - nRT in V2/V1 = - 2.303 nRT log V2/V1

GIVEN THAT :

• n ⟹ no. of molecules ➙ 1 mole.
• R ⟹ gas constant ➙ 2 cal.
• T ⟹ constant temperature associated with the process ➙ 25℃=(25+273)K=298K.
• V1 ⟹ Initial volume ➙ 15L
• V2 ⟹ Final volume ➙ 50L

➡ W = −2.303 × 1 × 2 × 298 × log (50/15)

➡ W = −1372.588 × (log 10 − log 3)

➡ W = −1372.588 × 0.523 = −717.86 ≈ = − 718 cal.

∴ The work done will be - 718 cal. (c)

Explanation:

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