P21_003 - E int = Q − W . Now the internal energy of an...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
3. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n ,thevo lume V ,and the temperature T by p = nRT/V . The work done by the gas during the isothermal expansion is W = Z V 2 V 1 pdV = nRT Z V 2 V 1 dV V = nRT ln V 2 V 1 . We substitute V 2 =2 V 1 to obtain W = nRT ln 2 = (4 . 00 mol) µ 8 . 31 J mol · K (400 K) ln2 = 9 . 22 × 10 3 J . (b) Since the expansion is isothermal, the change in entropy is given by ∆ S = R (1 /T ) dQ = Q/T , where Q
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: E int = Q − W . Now the internal energy of an ideal gas depends only on the temperature and not on the pressure and volume. Since the expansion is isothermal, ∆ E int = 0 and Q = W . Thus, ∆ S = W T = 9 . 22 × 10 3 J 400 K = 23 . 1 J / K . (c) ∆ S = 0 for all reversible adiabatic processes....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online