Michakato 30 ya thermodynamics ya Isobaric - matatizo na suluhisho
1. Mchoro wa PV hapa chini inaonyesha gesi bora hupitia isobariki mchakato. Hesabu kazi inafanywa na gesi katika mchakato AB.
Inajulikana:
Shinikizo (P) = 5 x 105 N / m2
Kiasi cha awali (V 1 ) = 2 m 3
Kiasi cha mwisho (V 2 ) = 6 m 3
Inayohitajika: Kazi (W)
Suluhisho:
W = P (V 2 – V 1 )
W = (5 x 105)(6 - 2) = (5 x 105) (4)
W = 20 x 105 = 2x106 Joule
2. Kuna tofauti gani kati ya kazi inayofanywa na gesi katika mchakato AB na mchakato CD…
Inajulikana:
Mchakato wa Isobaric AB :
Shinikizo (P) = 6 atm = 6 x 105 N / m2
Kiasi cha awali (V)1) = lita 1 = dm 13 = 1x10-3 m3
Juzuu ya mwisho (V)2) = lita 3 = dm 33 = 3x10-3 m3
CD ya mchakato wa Isobaric :
Shinikizo (P) = 4 atm = 4 x 105 N / m2
Kiasi cha awali (V 1 ) = lita 2 = 2 dm 3 = 2 x 10 -3 m 3
Kiasi cha mwisho (V 2 ) = lita 5 = 5 dm 3 = 5 x 10 -3 m 3
Alitaka : Tofauti ya kazi hufanywa na gesi katika mchakato AB na CD.
Suluhisho:
Kazi inafanywa na gesi katika mchakato AB:
W = P (V 2 – V 1 )
W = (6 x 10 5 )(3 x 10 -3 – 1 x 10 -3 )
W = (6 x 105)(2 x 10-3)
W = 12 x 102 = Jouli 1200
Kazi inafanywa na CD ya gesi katika mchakato:
W = P (V 2 – V 1 )
W = (4 x 10 5 )(5 x 10 -3 – 2 x 10 -3 )
W = (4 x 105)(3 x 10-3)
W = 12 x 102 = Jouli 1200
Tofauti ya kazi hufanywa na gesi katika mchakato AB na CD = 1200 - 1200 = 0.
3. Kazi inafanywa na gesi katika mchakato ABC ni….
Inajulikana:
Shinikizo 1 (P)1) = 6 x 105 Pa = 6 x 105 N / m2
Shinikizo 2 (P)2) = 3 x 105 Pa = 3 x 105 N / m2
Juzuu ya 1 (V1) = sm 23 = 2x10-6 m3
Juzuu ya 2 (V2) = sm 63 = 6x10-6 m3
Alitaka Kazi inafanywa katika mchakato wa ABC.
Suluhisho:
Katika mchakato AB, ujazo huwekwa sawa ili hakuna kazi inayofanywa na gesi.
Kazi ilifanywa na gesi katika mchakato wa BC.
W = P2 (V2 - V1)
W = (3 x 105)(6 x 10-6 - 2 x 10-6)
W = (3 x 105)(4 x 10-6)
W = 12 x 10-1
W = Jouli 1.2
Kazi inafanywa katika mchakato ABC = kazi inafanywa katika mchakato AB = 1.2 Jouli.
4. Amua mabadiliko katika nishati ya ndani kwa moles 2 za gesi bora inayopitia upanuzi wa isobariki kwa 300 K, ambapo \(\Delta V = 1\ \text{m}^3\).
Suluhisho: \(\Delta U = nC_v\Delta T\), kwa kutumia \(C_v = \frac{R}{\gamma-1}\) (kwa gesi bora ya monatomiki, \(\gamma = \frac{5}{3}\)) na \(\Delta T = \frac{P\Delta V}{nR}\), \(\Delta U = \frac{2\cdot 300 \cdot 1}{\frac{5}{3}-1} \takriban 1800\ \text{J}\).
5. Kokotoa uhamishaji wa joto katika mchakato wa isobariki ambapo mole 1 ya gesi bora ya diatomiki hupanuka, \(C_p = \frac{7}{2}R\), na \(\Delta T = 50\ \text{K}\).
Suluhisho: \(Q = nC_p\Delta T = \frac{7}{2} \cdot 50 \cdot R \takriban 1750\ \text{J}\) (kwa kutumia \(R = 8.314\ \text{J/(mol·K)}\)).
6. Tafuta kazi inayofanywa na mfumo unaopitia upanuzi wa isobaric, \(P = 3\ \text{atm}\), \(\Delta V = 4\ \text{L}\).
Suluhisho: \(W = P\Delta V = 3 \mara 4 = 12\ \maandishi{L·atm}\).
7. Amua mabadiliko katika entropi kwa mchakato wa isobariki ambapo molekuli 2 za gesi bora hubadilisha halijoto kwa 20 K. Tumia \(C_p = \frac{5}{2}R\).
Suluhisho: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 2 \cdot \frac{5}{2}R \cdot \ln\frac{T_1+20}{T_1}\).
8. Kokotoa uhamishaji wa joto kwa mgandamizo wa isobariki wa gesi bora ya monatomiki, \(C_p = \frac{5}{2}R\), \(\Delta T = -10\ \text{K}\).
Suluhisho: \(Q = nC_p\Delta T = \frac{5}{2} \cdot (-10) \cdot R \takriban -415\ \text{J}\).
9. Tafuta kazi iliyofanywa kwenye mfumo katika mchakato wa isobariki ukitumia \(P = 5\ \text{bar}\), \(\Delta V = -3\ \text{m}^3\).
Suluhisho: \(W = P\Delta V = 5 \mara (-3) = -15\ \text{bar m}^3\).
10. Amua mabadiliko katika nishati ya ndani kwa mchakato wa isobariki ambapo \(n = 3\ \text{mol}\), \(C_v = 3R\), \(\Delta T = 25\ \text{K}\).
Suluhisho: \(\Delta U = nC_v\Delta T = 3 \cdot 3R \cdot 25 \takriban 1883\ \text{J}\).
11. Kokotoa mabadiliko ya entropi katika mchakato wa isobariki kwa gesi bora ya diatomiki, \(n = 1\ \text{mol}\), \(\Delta T = 40\ \text{K}\), \(T_1 = 300\ \text{K}\).
Suluhisho: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = \frac{7}{2}R\ln\frac{340}{300}\).
12. Tafuta uhamisho wa joto katika upanuzi wa isobariki, \(P = 2\ \text{atm}\), \(\Delta V = 3\ \text{L}\), \(C_p = \frac{7}{2}R\).
Suluhisho: \(Q = P\Delta V + nC_p\Delta T = 2 \mara 3 + \frac{7}{2}R\Delta T\).
13. Amua kazi iliyofanywa katika mchakato wa isobaric kwa \(P = 4\ \text{bar}\), \(\Delta V = 5\ \text{m}^3\).
Suluhisho: \(W = P\Delta V = 4 \mara 5 = 20\ \text{bar m}^3\).
14. Kokotoa mabadiliko ya nishati ya ndani kwa mgandamizo wa isobariki, \(n = 2\ \text{mol}\), \(C_v = \frac{3}{2}R\), \(\Delta T = -30\ \text{K}\).
Suluhisho: \(\Delta U = nC_v\Delta T = 2 \cdot \frac{3}{2}R \cdot (-30) \takriban -753\ \text{J}\).
15. Tafuta mabadiliko ya entropi katika mchakato wa isobariki, \(n = 1.5\ \text{mol}\), \(\Delta T = 60\ \text{K}\), \(T_1 = 400\ \text{K}\), \(C_p = \frac{5}{2}R\).
Suluhisho: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 1.5 \cdot \frac{5}{2}R\ln\frac{460}{400}\).
16. Amua uhamishaji wa joto kwa upanuzi wa isobariki, \(P = 3\ \text{bar}\), \(\Delta V = 2\ \text{m}^3\), \(C_p = \frac{5}{2}R\), \(n = 2\ \text{mol}\).
Suluhisho: \(Q = P\Delta V + nC_p\Delta T = 3 \mara 2 + 2 \cdot \frac{5}{2}R\Delta T\).
17. Hesabu kazi iliyofanywa kwenye moles 3 za gesi inayopitia mgandamizo wa isobariki, \(P = 5\ \text{atm}\), \(\Delta V = -4\ \text{L}\).
Suluhisho: \(W = P\Delta V = 5 \mara (-4) = -20\ \maandishi{L·atm}\).
18. Tambua mabadiliko ya nishati ya ndani kwa \(n = 4\ \text{mol}\), \(C_v = \frac{7}{2}R\), \(\Delta T = 15\ \text{K}\) katika mchakato wa isobaric.
Suluhisho: \(\Delta U = nC_v\Delta T = 4 \cdot \frac{7}{2}R \cdot 15 \takriban 3157\ \text{J}\).
19. Tafuta uhamishaji wa joto katika mchakato wa isobaric, \(P = 4\ \text{atm}\), \(\Delta V = 5\ \text{L}\), \(n = 2\ \text{mol}\), \(C_p = \frac{5}{2}R\).
Suluhisho: \(Q = P\Delta V + nC_p\Delta T = 4 \mara 5 + 2 \cdot \frac{5}{2}R\Delta T\).
20. Amua kazi iliyofanywa katika mgandamizo wa isobariki, \(P = 7\ \text{bar}\), \(\Delta V = -2\ \text{m}^3\).
Suluhisho: \(W = P\Delta V = 7 \mara (-2) = -14\ \text{bar m}^3\).
21. Hesabu mabadiliko ya nishati ya ndani kwa moles 3 za gesi bora inayopitia mchakato wa isobariki, \(C_v = \frac{5}{2}R\), \(\Delta T = 20\ \text{K}\).
Suluhisho: \(\Delta U = nC_v\Delta T = 3 \cdot \frac{5}{2}R \cdot 20 \takriban 1256\ \text{J}\).
22. Tafuta mabadiliko ya entropi kwa upanuzi wa isobariki, \(n = 1\ \text{mol}\), \(C_p = \frac{7}{2}R\), \(\Delta T = 30\ \text{K}\), \(T_1 = 250\ \text{K}\).
Suluhisho: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = \frac{7}{2}R\ln\frac{280}{250}\).
23. Amua uhamishaji wa joto katika mchakato wa isobaric, \(P = 6\ \text{bar}\), \(\Delta V = 4\ \text{m}^3\), \(n = 3\ \text{mol}\), \(C_p = \frac{3}{2}R\).
Suluhisho: \(Q = P\Delta V + nC_p\Delta T = 6 \mara 4 + 3 \cdot \frac{3}{2}R\Delta T\).
24. Hesabu kazi iliyofanywa na mfumo katika upanuzi wa isobaric ukitumia \(P = 8\ \text{bar}\), \(\Delta V = 3\ \text{m}^3\).
Suluhisho: \(W = P\Delta V = 8 \mara 3 = 24\ \text{bar m}^3\).
25. Amua mabadiliko ya nishati ya ndani kwa mchakato wa isobariki ambapo \(n = 2\ \text{mol}\), \(C_v = \frac{7}{2}R\), \(\Delta T = -10\ \text{K}\).
Suluhisho: \(\Delta U = nC_v\Delta T = 2 \cdot \frac{7}{2}R \cdot (-10) \takriban -878\ \text{J}\).
26. Tafuta mabadiliko ya entropi kwa gesi bora ya diatomu katika mgandamizo wa isobariki, \(n = 1.5\ \text{mol}\), \(T_1 = 350\ \text{K}\), \(\Delta T = -40\ \text{K}\).
Suluhisho: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 1.5 \cdot \frac{7}{2}R\ln\frac{310}{350}\).
27. Amua uhamishaji wa joto kwa moles 2 za gesi inayopitia upanuzi wa isobariki, \(P = 5\ \text{bar}\), \(\Delta V = 6\ \text{m}^3\), \(C_p = \frac{5}{2}R\).
Suluhisho: \(Q = P\Delta V + nC_p\Delta T = 5 \mara 6 + 2 \cdot \frac{5}{2}R\Delta T\).
28. Hesabu kazi iliyofanywa kwenye mfumo katika mgandamizo wa isobariki ukitumia \(P = 9\ \text{atm}\), \(\Delta V = -3\ \text{L}\).
Suluhisho: \(W = P\Delta V = 9 \mara (-3) = -27\ \maandishi{L·atm}\).
29. Amua mabadiliko ya nishati ya ndani kwa moles 3 za gesi inayopitia mchakato wa isobariki, \(C_v = \frac{3}{2}R\), \(\Delta T = 15\ \text{K}\).
Suluhisho: \(\Delta U = nC_v\Delta T = 3 \cdot \frac{3}{2}R \cdot 15 \takriban 564\ \text{J}\).
30. Tafuta mabadiliko ya entropi katika upanuzi wa isobariki, \(n = 4\ \text{mol}\), \(C_p = \frac{5}{2}R\), \(\Delta T = 25\ \text{K}\), \(T_1 = 300\ \text{K}\).
Suluhisho: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 4 \cdot \frac{5}{2}R\ln\frac{325}{300}\).
