{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ch20 - Chapter 20 1(a Since the gas is ideal its pressure p...

This preview shows pages 1–4. Sign up to view the full content.

837 Chapter 20 1. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n , the volume V , and the temperature T by p = nRT / V . The work done by the gas during the isothermal expansion is 2 2 1 1 2 1 ln . V V V V dV V W p dV n RT n RT V V We substitute V 2 = 2.00 V 1 to obtain   3 = ln2.00 = 4.00 mol 8.31 J/mol K 400 K ln2.00 = 9.22 10 J. W n RT (b) Since the expansion is isothermal, the change in entropy is given by 1 S T dQ Q T , where Q is the heat absorbed. According to the first law of thermodynamics, 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, (c) S = 0 for all reversible adiabatic processes. 2. An isothermal process is one in which T i = T f , which implies ln ( T f / T i ) = 0. Therefore, Eq. 20-4 leads to 22.0 = ln = = 2.41 mol. 8.31 ln 3.9/1.3 f i V S nR n V 3. An isothermal process is one in which T i = T f , which implies ln( T f / T i ) = 0. Therefore, with V f / V i = 2.00, Eq. 20-4 leads to  = ln = 2.50 mol 8.31 J/mol K ln 2.00 =14.4 J/K. f i V S nR V 4. From Eq. 20-2, we obtain 3 9.22 10 J = = = 23.1 J/K. 400 K W S T

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
CHAPTER 20 838  4 = = 453 K 46.0 J/K = 2.08 10 J. Q T S 5. We use the following relation derived in Sample Problem — “Entropy change of two blocks coming to equilibrium:” = ln . f i T S mc T (a) The energy absorbed as heat is given by Eq. 19-14. Using Table 19-3, we find  4 J = = 386 2.00 kg 75 K = 5.79 10 J kg K Q cm T where we have used the fact that a change in Kelvin temperature is equivalent to a change in Celsius degrees. (b) With T f = 373.15 K and T i = 298.15 K, we obtain J 373.15 = 2.00 kg 386 ln =173 J/K. kg K 298.15 S 6. (a) This may be considered a reversible process (as well as isothermal), so we use S = Q / T where Q = Lm with L = 333 J/g from Table 19-4. Consequently,  333 J/g 15.0 g = =18.3 J/K. 273 K S (b) The situation is similar to that described in part (a), except with L = 2256 J/g, m = 5.00 g, and T = 373 K. We therefore find S = 30.2 J/K. 7. (a) We refer to the copper block as block 1 and the lead block as block 2. The equilibrium temperature T f satisfies m 1 c 1 ( T f T i ,1 ) + m 2 c 2 ( T f T i 2 ) = 0, which we solve for T f :       1 1 ,1 2 2 ,2 1 1 2 2 + 50.0 g 386 J/kg K 400 K + 100 g 128 J/kg K 200 K + 50.0 g 386 J/kg K + 100 g 128 J/kg K 320 K. i i f m c T m c T T m c m c (b) Since the two-block system in thermally insulated from the environment, the change in internal energy of the system is zero.
839 (c) The change in entropy is   1 2 1 1 2 2 ,1 ,2 = + = ln + ln 320 K 320 K = 50.0 g 386 J/kg K ln + 100 g 128 J/kg K ln 400 K 200 K 1.72 J K.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 38

ch20 - Chapter 20 1(a Since the gas is ideal its pressure p...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online