Irodov – Problems in General Physics

2.112. (a) dN= (2nno/a^3 /^2 ) e-UMT V U dU; (b) Up, =^1 /^2 kT.

2.113. In the latter case.

2.114. (a) II= 1 — ni--? = 0.25; (b) = 1 — ni /Y-1 = 0.18.

2.115. 8 = (1 — 11)/1 = 9.

2.116. -11 = 1 — 2T^3 /(T^1 + T^2 ).

2.117. n = — = 60%.

2.118. 1 1 = 1 — n-(1-1/^1 )•

2.119. 71 = 1 — (n + 7)41 + yn)

2.120. In both cases 11 =1

2.121. In both cases Ti =1 n

• 1

2.122. =1 : nn • 1 -"I

n - 1

2.123. (a) 11=1— 7 --‘, , (b) = 1^

ni — 1 y (n-1)^0 -1^

(n—i)

2.124. (a) 11= 1 n-1-1-(y — 1) n Inn '

(b) it =1

n-1+(y-1)Inn

y (n-1)

2.125. —

(T-1) In v

ti In v —1)/(y —1) '

2.126. rl =

(t— 1) In n

ti In n (T-1) y/(y —1) •

2.127. 11-1 2 7+1/T

(1-1-v) (1-1-1 77 )

2.128. The inequality .Q^1 — (^8) T2 Q; <0 becomes even stronger
when Ti is replaced by T,,,„„ and T^2 by Trnin Then Qiirmax
Q;amin< O. Hence
Qi—Q; Tnta:m—Turin a
<^ or 1<llcarnot• pdp^ (^61 41
2.129. According to the Carnot theorem (^) P
WW 1 = dTIT. Let us find the expressions
for 811 and 8Q 1. For an infinitesimal Carnot
cycle (e.g. parallelogram^1234 shown in
Fig. 14)
SA = dp-dV = (OpIOT)vdT • dV,

= p dV = [(away), + p] dV.^ Fig. 14.

It remains to substitute the two latter expressions into the former one.

2.130. (a) AS =

Ryln1.

19 JAK • mol); (b) AS — yR In y — 1 n =

= 25 J/(K • mol).

2.131. n = eAs/". = 2.0.

2.132. AS = vR ln n = 20 PK.

2.133. AS = M y-1 In n = —10 37K.

2.134. AS = In a — in 6) vR/(? — 1) = —11 J/K.

lkfc/V