Unformatted text preview: e final volume Vf . Solving Equation 15.5 for
3
the final volume yields Vfγ = PV γ
Vf = i i
P
f 1γ PViγ
i
, or
Pf
1γ P
= Vi i Pf ( = 6.34 × 10 −3 3 2.20 × 105 Pa 5
−2
3
m = 1.15 × 10 m
4 8.15 × 10 Pa 3 ) In the adiabatic process, the gas does an amount of work Wad given by Wad = 3 nR (Ti − Tf )
2 (15.4) In Equation 15.4, n is the number of moles of the gas, R is the universal gas constant, and Ti
and Tf are the initial and final Kelvin temperatures of the gas. In order to determine the
initial and final temperatures, we will employ the ideal gas law: PV = nRT (Equation 14.1).
Solving Equation 14.1 for the initial and final Kelvin temperatures yields 784 THERMODYNAMICS Ti = PiVi and nR Tf = Pf Vf (3) nR SOLUTION Substituting Equations (3) into Equation 15.4 gives
Wad = 3 nR (Ti − Tf ) =
2 3
2 PV P V nR i i − f f =
nR nR 3
2 ( PVi − Pf Vf )
i (4) Substituting Equations (4) and (2) into Equation (1), we obtain
Wdiff = Wad − Walt = 3
2 ( PVi − Pf Vf ) − Pf (Vf − Vi )
i = 3 PV
2ii 3
3
5
− 2 Pf Vf − Pf Vf + Pf Vi = 2 PVi − 2 Pf Vf + Pf Vi
i = 3 PV
2ii + Pf Vi − 5 Vf
2 ( ) Substituting in the given values of Pi , Pf , Vi , and the calculated value of Vf , yields
Wdiff = 3
2 5
( 2.20 × 105 Pa ) ( 6.34 × 10−3 m 3 ) + (8.15 × 104 Pa ) 6.34 × 10−3 m 3 − 2 (1.15 × 10−2 m 3 ) = 270 J 33. SSM REASONING AND SOLUTION Let the left be side 1 and the right be side 2. Since
the partition moves to the right, side 1 does work on side 2, so that the work values involved
satisfy the relation W1 = −W2 . Using Equation 15.4 for each work value, we find that
3
nR
2 (T1i –T1f ) = − 3 nR (T2i –T2f )
2 or T1f + T2f = T1i + T2i = 525 K + 275 K = 8.00 × 102 K
We now seek a second equation for the two unknowns T1f and T2f . Equation 15.5 for an
γ
γ
γ
γ
adiabatic process indicates that P iV1i = P V1f and P2iV2i = P2f V2f . Dividing these two
1
1f equations and using the facts that V1i = V2i and P = P2f , gives
1f
γ
P iV1i
1 γ
P2iV2i = γ
P V1f
1f γ
P2f V2f Using the ideal gas law, we find that or V
= 1f
P2i V2f P
1i γ Chapter 15 Problems V
= 1f
P2i V2f P
1i γ nRT1f /P
1f
= nRT /P
nRT2i /V2i 2f 2f
nRT1i /V1i becomes 785 γ Since V1i = V2i and P = P2f , the result above reduces to
1f T
= 1f
T2i T2f T1i γ or T1f
T2f 1/ γ T = 1i T 2i 1/ γ 525 K = 275 K = 1.474 Using this expression for the ratio of the final temperatures in T1f + T2f = 8.00 × 10 2 K , we
find that
a. T1f = 477 K and b. T2f = 323 K 34. REASONING AND SOLUTION
a. The amount of heat needed to raise the temperature of the gas at constant volume is given
by Equations 15.6 and 15.8, Q = n CV ∆T. Solving for ∆T yields
∆T = Q
5.24 × 103 J
=
= 1.40 × 102 K
nCv ( 3.00 mol ) 3 R
2 () b. The change in the internal energy of the gas is given by the first law of thermodynamics
with W = 0, since the gas is heated at constant volume:
∆U = Q − W = 5.24 × 103 J − 0 = 5.24 × 103 J c. The change in pressure can be obtained from the ideal gas law, ( ) 2
nR ∆T ( 3.00 mol ) R 1.40 × 10 K
∆P =
=
= 2.33 × 103 Pa
V
1.50 m3 35. SSM REASONING AND SOLUTION According to the first law of thermodynamics
(Equation 15.1), ∆U = U f − U i = Q − W . Since the internal energy of this gas is doubled by
the addition of heat, the initial and final internal energies are U and 2U, respectively.
Therefore,
∆U = U f − U i = 2U − U = U
Equation 15.1 for this situation then becomes U = Q − W . Solving for Q gives 786 THERMODYNAMICS Q = U +W (1) The initial internal energy of the gas can be calculated from Equation 14.7:
3
2 U = nRT = 3
2 ( 2.5 mol ) 8.31 J/ ( mol ⋅ K ) ( 350 K ) = 1.1×104 J a. If the process is carried out isochorically (i.e., at constant volume), then W = 0, and the
heat required to double the internal energy is
Q = U + W = U + 0 = 1.1× 104 J b. If the process is carried out isobarically (i.e., at constant pressure), then W = P ∆V , and
Equation (1) above becomes
(2)
Q = U + W = U + P ∆V
From the ideal gas law, PV = nRT , we have that P ∆V = nR ∆T , and Equation (2) becomes Q = U + nR ∆T (3) The internal energy of an ideal gas is directly proportional to its Kelvin temperature. Since
the internal energy of the gas is doubled, the final Kelvin temperature will be twice the
initial Kelvin temperature, or ∆T = 350 K. Substituting values into Equation (3) gives
Q = 1.1× 10 4 J + (2.5 mol)[8.31 J/(mol ⋅ K)](350 K) = 1.8 × 104 J 36. REASONING The gas is in a rigid container, so we conclude that the heating takes place at
constant volume. For a constantvolume process involving an ideal monatomic gas, the
amount Q of heat transferred is given by Q = CV n ∆T (Equation 15.6), where C V = 3 R
2
(Equation 15.8) is the molar specific heat capacity at constant volume, R is the universal gas
constant, n is the number of moles of gas, and ∆T is the change in temperature.
SOLUTION Solving Q = CV n ∆T (Equation 15.6) for the number n of moles of the gas,
we obtain
Q
(1)
n=
C V ∆T
Substituting C V = 3 R (E...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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