30 Izotermik termodinamik jarayonlar – muammolar va yechimlar
1. Quyidagi PV diagrammasida ideal gaz izotermik jarayondan o'tishi ko'rsatilgan . AB jarayonida gaz tomonidan bajarilgan ishni hisoblang.
qaror
ish gaz tomonidan bajarilgan ish PV egri chizig'i ostidagi maydonga teng
AB = uchburchak maydoni + to'rtburchak maydoni
W = [½ (8 x 10 5 –4 x 10 5 )(3-1)] + [4 x 10 5 (3-1)]
W = [½ (4 x 10 5 )(2)] + [4 x 10 5 (2)]
V = [4 x 10 5 ] + [8 x 10 5 ]
W = 12 x 105 Joules
AB = 12 x 10 5 Joul jarayonida gaz tomonidan ish bajariladi
2. ABC jarayonida ideal gaz tomonidan bajarilgan ishni hisoblang.
Ish ideal gaz tomonidan ABC = jarayonida bajariladi. PV egri chizig'i ostidagi maydon
AB = uchburchak maydoni + to'rtburchak maydoni
Vt = [½(10×10 5 –5×10 5 )(30-10)]+[5×10 5 (30-10)]
W = [½ (5 x 10 5 )(20)] + [5 x 10 5 (20)]
W = [(5 x 10 5 )(10)] + [100 x 10 5 ]
V = [50 x 10 5 ] + [100 x 10 5 ]
W = 150 x 105 Joules
W = 1.5 x 105 Joules
3. Izotermik jarayonlardan o'tayotgan ideal gaz. Gazga qancha issiqlik qo'shiladi, shunda gaz atrof-muhitga 5000 Joul miqdorida ishlaydi.
Ma'lum:
Ish (V) = 5000 Joul
Istalgan: Gazga issiqlik qo'shiladi (Q)
yechim:
Izotermik jarayon - bu doimiy haroratda sodir bo'ladigan termodinamik jarayon.
ΔU = 3/2 n R ΔT
ΔU = ichki energiyaning o'zgarishi, n = mollar soni, R = universal gaz doimiysi, ΔT = Haroratning o'zgarishi.
Yuqoridagi tenglamaga ko'ra, agar ΔT = 0 bo'lsa, u holda ΔU = 0 bo'ladi.
Termodinamikaning birinchi qonunining tenglamasi:
ΔU = Q – W
0 = Q – W
Q = W
Q = 5000 Joul.
4. Izotermik jarayondan o'tayotgan ideal gaz uchun foton nurlanish diagrammasi quyidagi rasmda ko'rsatilgan. AB jarayonidagi gaz tomonidan qo'shiladigan issiqlikni hisoblang.
Ma'lum:
Bosim 1 (P1 ) = 5 atm = 5 x 10 5 Pa
Bosim 2 (P2 ) = 10 atm = 10 x 10 5 Pa
1-hajm (V1 ) = 2 m3
2-hajm (V2 ) = 6 m3
Kerakli : AB jarayonida issiqlik qo'shiladi.
yechim:
Izotermik = doimiy harorat. Quyidagi tenglamaga ko'ra, agar ΔT = 0 bo'lsa, ΔU = 0.
ΔU = 3/2 n R ΔT
ΔU = 3/2 n R (0)
ΔU = 0
Termodinamikaning birinchi qonuniga qo'llang :
ΔU = QW
0 = QW
Q=W
Ish ideal gaz tomonidan bajariladi = foton egri chizig'i ostidagi maydon = uchburchak maydoni + to'rtburchak maydoni
W = ½ (P2 – P1 ) (V2 – V1 ) + P1 ( V2 – V1 )
W = ½ (10 x 10 5 – 5 x 10 5 )(6-2) + (5 x 10 5 )(6-2)
W = ½ (5 x 10 5 )(4) + (5 x 10 5 )(4)
W = ½ (20 x 10 5 ) + (20 x 10 5 )
V = (10 x 10 5 ) + (20 x 10 5 )
W = 30 x 10 5 Joul
5. 300 K haroratda 2 l dan 4 l gacha izotermik ravishda kengaygan 1 mol ideal gazda bajarilgan ishni hisoblang.
Yechim: \( W = nRT \ln\left(\frac{V_2}{V_1}\right) = 1 \times 8.314 \times 300 \times \ln(2) \approx 1724 \, \text{J} \)
6. Yuqoridagi masala uchun issiqlik uzatishni aniqlang.
Yechim: \( Q = W = 1724 \, \text{J} \)
7. Yuqoridagi jarayon uchun ichki energiyaning o'zgarishini hisoblang.
Yechim: \( \Delta U = 0 \, \text{J} \) (chunki jarayon izotermik)
8. Ideal gazni 300 K haroratda 10 L dan 5 L gacha izotermik siqish uchun bajarilgan ishni toping.
Yechim: \( W = 8.314 \marta 300 \marta \ln\left(\frac{5}{10}\right) \approx -862 \, \text{J} \)
9. 1-masaladagi izotermik kengayish uchun entropiya o'zgarishini hisoblang.
Yechim: \( \Delta S = nR\ln\left(\frac{V_2}{V_1}\right) = 8.314 \times \ln(2) \approx 5.76 \, \text{J/K} \)
10. 200 K haroratda 2 mol ideal gazning 4 l dan 2 l gacha izotermik siqilishi uchun issiqlik uzatishni toping.
Yechim: \( Q = 2 \times 8.314 \times 200 \times \ln\left(\frac{2}{4}\right) \approx -1152 \, \text{J} \)
11. Izotermik jarayon uchun Gibbsning erkin energiyasining o'zgarishini hisoblang.
Yechim: \( \Delta G = 0 \) (Yopiq tizimda qaytar izotermik jarayon uchun, \( \Delta G = 0 \))
12. 400 K haroratda 3 mol ideal gazning 6 l dan 3 l gacha izotermik siqilishi uchun entropiya o'zgarishini aniqlang.
Yechim: \( \Delta S = 3 \times 8.314 \times \ln\left(\frac{3}{6}\right) \approx -17.29 \, \text{J/K} \)
13. Ideal gaz 250 K da 2 mol davomida izotermik ravishda 3 L dan 6 L gacha kengayganda bajarilgan ishni toping.
Yechim: \( W = 2 \times 8.314 \times 250 \times \ln(2) \approx 2874 \, \text{J} \)
14. Yuqoridagi jarayon uchun issiqlik uzatishni aniqlang.
Yechim: \( Q = W = 2874 \, \text{J} \)
15. Yuqoridagi izotermik kengayish uchun ichki energiyaning o'zgarishini hisoblang.
Yechim: \( \Delta U = 0 \, \text{J} \)
16. 500 K haroratda 4 mol ideal gazning 5 l dan 10 l gacha izotermik kengayishi uchun entropiya o'zgarishini hisoblang.
Yechim: \( \Delta S = 4 \marta 8.314 \marta \ln(2) \taxminan 23.03 \, \text{J/K} \)
17. 300 K haroratda 8 L dan 4 L gacha bo'lgan 1 mol ideal gazning izotermik siqilishi uchun issiqlik uzatishni aniqlang.
Yechim: \( Q = 8.314 \marta 300 \marta \ln\left(\frac{4}{8}\right) \approx -862 \, \text{J} \)
18. 350 K haroratda 3 mol ideal gaz izotermik ravishda 3 l dan 9 l gacha kengayganda bajarilgan ishni toping.
Yechim: \( W = 3 \times 8.314 \times 350 \times \ln(3) \approx 5362 \, \text{J} \)
19. Yuqoridagi jarayon uchun entropiya o'zgarishini hisoblang.
Yechim: \( \Delta S = 3 \marta 8.314 \marta \ln(3) \taxminan 14.88 \, \text{J/K} \)
20. 200 K haroratda 2 mol ideal gazning 2 l dan 8 l gacha izotermik kengayishi uchun issiqlik uzatishni aniqlang.
Yechim: \( Q = 2 \marta 8.314 \marta 200 \marta \ln(4) \taxminan 2304 \, \text{J} \)
21. 500 K haroratda 4 mol ideal gaz 10 L dan 5 L gacha izotermik siqilish uchun ichki energiyaning o'zgarishini hisoblang.
Yechim: \( \Delta U = 0 \, \text{J} \)
22. 300 K haroratda 5 mol ideal gazning 5 l dan 15 l gacha izotermik kengayishi uchun entropiya o'zgarishini aniqlang.
Yechim: \( \Delta S = 5 \marta 8.314 \marta \ln(3) \taxminan 24.81 \, \text{J/K} \)
23. 2 mol ideal gaz 400 K haroratda 6 l dan 2 l gacha izotermik ravishda siqilganda bajarilgan ishni toping.
Yechim: \( W = 2 \times 8.314 \times 400 \times \ln\left(\frac{2}{6}\right) \approx -2874 \, \text{J} \)
24. 250 K haroratda 3 mol ideal gazning 9 L dan 3 L gacha izotermik siqilishi uchun entropiya o'zgarishini hisoblang.
Yechim: \( \Delta S = 3 \times 8.314 \times \ln\left(\frac{3}{9}\right) \approx -14.88 \, \text{J/K} \)
25. 350 K haroratda 1 mol ideal gazning 4 l dan 12 l gacha izotermik kengayishi uchun issiqlik uzatishni aniqlang.
Yechim: \( Q = 8.314 \marta 350 \marta \ln(3) \taxminan 1791 \, \text{J} \)
26. 500 K haroratda 4 mol ideal gaz izotermik ravishda 4 l dan 8 l gacha kengayganda bajarilgan ishni toping.
Yechim: \( W = 4 \times 8.314 \times 500 \times \ln(2) \approx 5749 \, \text{J} \)
27. 200 K haroratda 10 L dan 5 L gacha bo'lgan 2 mol ideal gazning izotermik siqilishi uchun entropiya o'zgarishini aniqlang.
Yechim: \( \Delta S = 2 \times 8.314 \times \ln\left(\frac{5}{10}\right) \approx -5.76 \, \text{J/K} \)
24. 3 mol ideal gaz 450 K haroratda 9 l dan 3 l gacha izotermik ravishda siqilganda bajarilgan ishni toping.
Yechim: \( W = 3 \times 8.314 \times 450 \times \ln\left(\frac{3}{9}\right) \approx -4310 \, \text{J} \)
25. 400 K haroratda 5 mol ideal gazning 5 l dan 10 l gacha izotermik kengayishi uchun ichki energiyaning o'zgarishini hisoblang.
Yechim: \( \Delta U = 0 \, \text{J} \)
26. 300 K haroratda 4 mol ideal gazning 6 l dan 18 l gacha izotermik kengayishi uchun entropiya o'zgarishini aniqlang.
Yechim: \( \Delta S = 4 \marta 8.314 \marta \ln(3) \taxminan 19.85 \, \text{J/K} \)
Bu muammolar izotermik jarayonlarning turli jihatlarini, jumladan, bajarilgan ish, issiqlik uzatish, ichki energiyaning o'zgarishi va entropiyaning o'zgarishini qamrab oladi.
