30 ta izobarik termodinamika jarayonlari – muammolar va yechimlar
1. Quyidagi PV diagrammasida ideal gaz izobarik jarayondan o'tishi ko'rsatilgan . AB jarayonida gaz tomonidan bajarilgan ishni hisoblang.
Ma'lum:
Bosim (P) = 5 x 10 5 N/m 2
Boshlang'ich hajm (V1 ) = 2 m3
Yakuniy hajm (V2 ) = 6 m3
Qidirilmoqda: Ish (W)
yechim:
W = P ( V2 – V1 )
W = (5 x 10 5 )(6 – 2) = (5 x 10 5 ) (4)
W = 20 x 10 5 = 2 x 10 6 Joules
2. AB va CD jarayonlarida gaz tomonidan bajarilgan ish o'rtasida qanday farq bor...
Ma'lum:
Izobarik jarayon AB :
Bosim (P) = 6 atm = 6 x 10 5 N/m 2
Boshlang'ich hajm (V1 ) = 1 litr = 1 dm3 = 1 x 10 -3 m3
Yakuniy hajm (V2 ) = 3 litr = 3 dm3 = 3 x 10 -3 m3
Izobarik jarayon CD :
Bosim (P) = 4 atm = 4 x 10 5 N/m 2
Boshlang'ich hajm (V1 ) = 2 litr = 2 dm3 = 2 x 10 -3 m3
Yakuniy hajm (V2 ) = 5 litr = 5 dm3 = 5 x 10 -3 m3
Kerakli : AB va CD jarayonlaridagi gaz tomonidan bajarilgan ishning farqi.
yechim:
AB jarayonidagi gaz tomonidan ish bajariladi:
W = P ( V2 – V1 )
W = (6 x 10 5 ) (3 x 10 -3 – 1 x 10 -3 )
W = (6 x 10 5 )(2 x 10 -3 )
W = 12 x 10 2 = 1200 Joul
Ish jarayondagi CD gazi tomonidan bajariladi:
W = P ( V2 – V1 )
W = (4 x 10 5 ) (5 x 10 -3 – 2 x 10 -3 )
W = (4 x 10 5 )(3 x 10 -3 )
W = 12 x 10 2 = 1200 Joul
Ishning farqi AB va CD jarayonidagi gaz tomonidan bajariladi = 1200 – 1200 = 0.
3. Jarayondagi gaz tomonidan ish bajariladi. ABC….
Ma'lum:
Bosim 1 (P1 ) = 6 x 10 5 Pa = 6 x 10 5 N/ m2
Bosim 2 (P2 ) = 3 x 10 5 Pa = 3 x 10 5 N/ m2
1 -hajm (V1 ) = 2 sm3 = 2 x 10 -6 m3
2 -hajm (V2 ) = 6 sm3 = 6 x 10 -6 m3
Qidirilmoqda : Ish ABC jarayonida bajarilmoqda.
yechim:
AB jarayonida gaz tomonidan hech qanday ish bajarilmasligi uchun hajm doimiy ravishda saqlanadi.
Miloddan avvalgi jarayonda gaz tomonidan ish bajarilgan.
W = P2 ( V2 – V1 )
W = (3 x 10 5 )(6 x 10 -6 – 2 x 10 -6 )
W = (3 x 10 5 )(4 x 10 -6 )
Vt = 12 x 10 -1
W = 1.2 Joul
Ish ABC jarayonida bajariladi = ish AB = 1.2 Joul jarayonida bajariladi.
4. 300 K da izobarik kengayishga uchragan ideal gazning 2 molining ichki energiya o'zgarishini aniqlang, bu yerda \(\Delta V = 1\ \text{m}^3\).
Yechim: \(\Delta U = nC_v\Delta T\), \(C_v = \frac{R}{\gamma-1}\) (monatomik ideal gaz uchun, \(\gamma = \frac{5}{3}\)) va \(\Delta T = \frac{P\Delta V}{nR}\), \(\Delta U = \frac{2\cdot 300 \cdot 1}{\frac{5}{3}-1} \approx 1800\ \text{J}\) yordamida.
5. 1 mol diatomik ideal gaz kengayadigan izobarik jarayonda issiqlik uzatishni hisoblang, \(C_p = \frac{7}{2}R\) va \(\Delta T = 50\ \text{K}\).
Yechim: \(Q = nC_p\Delta T = \frac{7}{2} \cdot 50 \cdot R \approx 1750\ \text{J}\) (\(R = 8.314\ \text{J/(mol·K)}\) yordamida).
6. Izobar kengayishni boshdan kechirayotgan tizim tomonidan bajarilgan ishni toping, \(P = 3\ \text{atm}\), \(\Delta V = 4\ \text{L}\).
Yechim: \(W = P\Delta V = 3 \times 4 = 12\ \text{L·atm}\).
7. Ideal gazning 2 mol harorati 20 K ga o'zgarganda izobarik jarayon uchun entropiyaning o'zgarishini aniqlang. \(C_p = \frac{5}{2}R\) dan foydalaning.
Yechim: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 2 \cdot \frac{5}{2}R \cdot \ln\frac{T_1+20}{T_1}\).
8. Monatomik ideal gazning izobarik siqilishi uchun issiqlik uzatishni hisoblang, \(C_p = \frac{5}{2}R\), \(\Delta T = -10\ \text{K}\).
Yechim: \(Q = nC_p\Delta T = \frac{5}{2} \cdot (-10) \cdot R \approx -415\ \text{J}\).
9. \(P = 5\ \text{bar}\), \(\Delta V = -3\ \text{m}^3\) bo'lgan izobarik jarayonda tizimda bajarilgan ishni toping.
Yechim: \(W = P\Delta V = 5 \times (-3) = -15\ \text{bar m}^3\).
10. Izobarik jarayon uchun ichki energiyaning o'zgarishini aniqlang, bunda \(n = 3\ \text{mol}\), \(C_v = 3R\), \(\Delta T = 25\ \text{K}\).
Yechim: \(\Delta U = nC_v\Delta T = 3 \cdot 3R \cdot 25 \approx 1883\ \text{J}\).
11. Ikki atomli ideal gaz uchun izobarik jarayonda entropiya o'zgarishini hisoblang, \(n = 1\ \text{mol}\), \(\Delta T = 40\ \text{K}\), \(T_1 = 300\ \text{K}\).
Yechim: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = \frac{7}{2}R\ln\frac{340}{300}\).
12. Izobar kengayishdagi issiqlik uzatishni toping, \(P = 2\ \text{atm}\), \(\Delta V = 3\ \text{L}\), \(C_p = \frac{7}{2}R\).
Yechim: \(Q = P\Delta V + nC_p\Delta T = 2 \times 3 + \frac{7}{2}R\Delta T\).
13. \(P = 4\ \text{bar}\), \(\Delta V = 5\ \text{m}^3\) uchun izobarik jarayonda bajarilgan ishni aniqlang.
Yechim: \(W = P\Delta V = 4 \times 5 = 20\ \text{bar m}^3\).
14. Izobarik siqilish uchun ichki energiya o'zgarishini hisoblang, \(n = 2\ \text{mol}\), \(C_v = \frac{3}{2}R\), \(\Delta T = -30\ \text{K}\).
Yechim: \(\Delta U = nC_v\Delta T = 2 \cdot \frac{3}{2}R \cdot (-30) \approx -753\ \text{J}\).
15. Izobarik jarayonda entropiya o'zgarishini toping, \(n = 1.5\ \text{mol}\), \(\Delta T = 60\ \text{K}\), \(T_1 = 400\ \text{K}\), \(C_p = \frac{5}{2}R\).
Yechim: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 1.5 \cdot \frac{5}{2}R\ln\frac{460}{400}\).
16. Izobar kengayish uchun issiqlik uzatishni aniqlang, \(P = 3\ \text{bar}\), \(\Delta V = 2\ \text{m}^3\), \(C_p = \frac{5}{2}R\), \(n = 2\ \text{mol}\).
Yechim: \(Q = P\Delta V + nC_p\Delta T = 3 \times 2 + 2 \cdot \frac{5}{2}R\Delta T\).
17. Izobarik siqilishga uchragan gazning 3 molida bajarilgan ishni hisoblang, \(P = 5\ \text{atm}\), \(\Delta V = -4\ \text{L}\).
Yechim: \(W = P\Delta V = 5 \times (-4) = -20\ \text{L·atm}\).
18. Izobarik jarayonda \(n = 4\ \text{mol}\), \(C_v = \frac{7}{2}R\), \(\Delta T = 15\ \text{K}\) uchun ichki energiya o'zgarishini aniqlang.
Yechim: \(\Delta U = nC_v\Delta T = 4 \cdot \frac{7}{2}R \cdot 15 \approx 3157\ \text{J}\).
19. Izobarik jarayonda issiqlik uzatishni toping, \(P = 4\ \text{atm}\), \(\Delta V = 5\ \text{L}\), \(n = 2\ \text{mol}\), \(C_p = \frac{5}{2}R\).
Yechim: \(Q = P\Delta V + nC_p\Delta T = 4 \times 5 + 2 \cdot \frac{5}{2}R\Delta T\).
20. Izobarik siqilishda bajarilgan ishni aniqlang, \(P = 7\ \text{bar}\), \(\Delta V = -2\ \text{m}^3\).
Yechim: \(W = P\Delta V = 7 \times (-2) = -14\ \text{bar m}^3\).
21. Izobarik jarayondan o'tayotgan ideal gazning 3 mol hajmi uchun ichki energiya o'zgarishini hisoblang, \(C_v = \frac{5}{2}R\), \(\Delta T = 20\ \text{K}\).
Yechim: \(\Delta U = nC_v\Delta T = 3 \cdot \frac{5}{2}R \cdot 20 \approx 1256\ \text{J}\).
22. Izobarik kengayish uchun entropiya o'zgarishini toping, \(n = 1\ \text{mol}\), \(C_p = \frac{7}{2}R\), \(\Delta T = 30\ \text{K}\), \(T_1 = 250\ \text{K}\).
Yechim: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = \frac{7}{2}R\ln\frac{280}{250}\).
23. Izobar jarayonda issiqlik uzatishni aniqlang, \(P = 6\ \text{bar}\), \(\Delta V = 4\ \text{m}^3\), \(n = 3\ \text{mol}\), \(C_p = \frac{3}{2}R\).
Yechim: \(Q = P\Delta V + nC_p\Delta T = 6 \times 4 + 3 \cdot \frac{3}{2}R\Delta T\).
24. \(P = 8\ \text{bar}\), \(\Delta V = 3\ \text{m}^3\) bo'lgan izobarik kengayishda tizim tomonidan bajarilgan ishni hisoblang.
Yechim: \(W = P\Delta V = 8 \times 3 = 24\ \text{bar m}^3\).
25. Izobarik jarayon uchun ichki energiya o'zgarishini aniqlang, bu yerda \(n = 2\ \text{mol}\), \(C_v = \frac{7}{2}R\), \(\Delta T = -10\ \text{K}\).
Yechim: \(\Delta U = nC_v\Delta T = 2 \cdot \frac{7}{2}R \cdot (-10) \approx -878\ \text{J}\).
26. Izobarik siqilishdagi diatomik ideal gaz uchun entropiya o'zgarishini toping, \(n = 1.5\ \text{mol}\), \(T_1 = 350\ \text{K}\), \(\Delta T = -40\ \text{K}\).
Yechim: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 1.5 \cdot \frac{7}{2}R\ln\frac{310}{350}\).
27. Izobarik kengayishdan o'tayotgan 2 mol gaz uchun issiqlik uzatishni aniqlang, \(P = 5\ \text{bar}\), \(\Delta V = 6\ \text{m}^3\), \(C_p = \frac{5}{2}R\).
Yechim: \(Q = P\Delta V + nC_p\Delta T = 5 \times 6 + 2 \cdot \frac{5}{2}R\Delta T\).
28. \(P = 9\ \text{atm}\), \(\Delta V = -3\ \text{L}\) bo'lgan izobarik siqilishda tizimda bajarilgan ishni hisoblang.
Yechim: \(W = P\Delta V = 9 \times (-3) = -27\ \text{L·atm}\).
29. Izobarik jarayondan o'tayotgan 3 mol gaz uchun ichki energiya o'zgarishini aniqlang, \(C_v = \frac{3}{2}R\), \(\Delta T = 15\ \text{K}\).
Yechim: \(\Delta U = nC_v\Delta T = 3 \cdot \frac{3}{2}R \cdot 15 \approx 564\ \text{J}\).
30. Izobarik kengayishdagi entropiya o'zgarishini toping, \(n = 4\ \text{mol}\), \(C_p = \frac{5}{2}R\), \(\Delta T = 25\ \text{K}\), \(T_1 = 300\ \text{K}\).
Yechim: \(\Delta S = nC_p\ln\frac{T_2}{T_1} = 4 \cdot \frac{5}{2}R\ln\frac{325}{300}\).
