30 Termodinamikaning birinchi qonunining ba'zi termodinamik jarayonlarda qo'llanilishi (Izobarik Izotermik Izoxorik)
1. Quyidagi grafik gaz tomonidan boshdan kechiriladigan termodinamik siklni ko'rsatadi. ABCD jarayonida gaz tomonidan bajarilgan ish ...
Ma'lum:
Bosim 1 (P1 ) = 2 x 10 5 Pa
Bosim 2 (P2 ) = 4 x 10 5 Pa
1-hajm (V1 ) = 1 m3
2-hajm (V2 ) = 3 m3
Kerakli: ABCD (W) jarayonida gaz bilan bajarilgan ish.
yechim:
Gaz tomonidan bajarilgan ish ABCD maydoniga teng.
W = (P2 – P1 ) ( V2 – V1 )
W = (4 x 10 5 – 2 x 10 5 )(3 – 1)
W = (2 x 10 5 )(2)
W = 4 x 105 Joules
2. ABC jarayonida gazning bajaradigan ishi…
Ma'lum:
Bosim 1 (P1 ) = 3 x 10 5 Pa
Bosim 2 (P2 ) = 6 x 10 5 Pa
1 -hajm (V1 ) = 20 sm3 = 20 x 10 -6 m3
2 -hajm (V2 ) = 60 sm3 = 60 x 10 -6 m3
Izlanmoqda : ABC jarayonida gaz tomonidan bajarilgan ish
yechim:
Gaz tomonidan bajarilgan ish = ABC maydoni.
W = ½ (P2 – P1 ) ( V2 – V1 )
Vt = 1/2 (6 x 10 5 – 3 x 10 5 ) (60 x 10 -6 – 20 x 10 -6 )
Vt = 1/2 (3 x 10 5 ) (40 x 10 -6 )
W = 1/2 (120 x 10 -1 Joules)
W = 1/2 (12 Joules)
W = 6 Joul
3. Quyidagi rasmda yopiq idishdagi gaz uchun fotovoltaik kuchlanish diagrammasi ko'rsatilgan. Gaz tomonidan bajarilgan ish qaysi jarayonda ko'rsatilgan.
qaror
AB va DC jarayonlari izobarik jarayonlardir ( bosim doimiy ravishda saqlanadi) . AD va BC jarayonlari juda xor jarayonlardir ( hajm o'zgarmaydi ).
Gaz kengayganda ish gaz tomonidan bajarildi (jarayon DC).
Izobarik jarayon (doimiy bosim) bilan bog'liq 9 ta muammo va yechimlar:
Muammo 1:
Doimiy bosimdagi 3.0 m³ gaz 2.0 m³ gacha siqiladi. Agar boshlang'ich harorat 300 K bo'lsa, yakuniy harorat qanday bo'ladi?
Qarori:
Izobarik jarayon uchun tenglamadan foydalanib: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), biz oxirgi haroratni quyidagicha yechishimiz mumkin:
\[
\frac{3.0}{300} = \frac{2.0}{T_2} \T_2 = \frac{2.0 \times 300}{3.0} = 200\, \text{K} ni anglatadi
\]
Muammo 2:
2 atmosferada 1 litrdan 3 litrgacha izobarik kengaygan 2 mol gazning bajargan ishini hisoblang.
Qarori:
Izobarik jarayonda bajarilgan ish uchun tenglamadan foydalanish: \( W = P \Delta V \):
\[
W = 2\, \text{atm} \times (3 – 1)\, \text{litr} = 4\, \text{atm}\cdot\text{litr}
\]
Muammo 3:
4 mol ideal diatomik gaz 200 K dan 300 K gacha doimiy bosimda qizdirilganda ichki energiyaning o'zgarishi qanday bo'ladi?
Qarori:
Ikki atomli gaz uchun doimiy bosimdagi ichki energiyaning o'zgarishi tenglamasidan foydalanish: \( \Delta U = nC_pdT \), bu yerda \(C_p = \frac{7}{2}R \):
\[
\Delta U = 4 \times \frac{7}{2} \times 8.314 \times (300 – 200) \taxminan 6990\, \text{J}
\]
Muammo 4:
400 K haroratda 2 mol monoatomli gaz 5 litrdan 10 litrgacha kengayganda izobarik jarayon davomida qo'shilgan issiqlikni hisoblang.
Qarori:
Izobar jarayonda issiqlik uzatish tenglamasidan foydalanish: \( Q = nC_pdT \), bu yerda \( C_p = \frac{5}{2}R \) va \( \frac{V_2}{T_2} = \frac{V_1}{T_1} \):
\[
Q = 2 marta \frac{5}{2} \times 8.314 \times \left(400 \times \frac{10}{5} – 400\right) = 3316\, \text{J}
\]
Muammo 5:
200 K haroratda 1 litrdan 2 litrgacha kengaygan 3 mol triatomik gazning entalpiyasining o'zgarishini aniqlang.
Qarori:
Izobarik jarayon uchun entalpiyaning o'zgarishi formulasidan foydalanish: \( \Delta H = nC_pdT \), bu yerda \( C_p = \frac{f}{2}R \) va \( f = 6 \):
\[
\Delta H = 3 \times \frac{6}{2} \times 8.314 \times \left(200 \times \frac{2}{1} – 200\right) = 4989\, \text{J}
\]
Muammo 6:
300 K haroratda va dastlab 5 litr hajmdagi 2 mol gaz doimiy bosimda 500 K gacha qizdirilganda uning yakuniy hajmini hisoblang.
Qarori:
Izobarik jarayon uchun tenglamadan foydalanish:
\[
\frac{V_1}{T_1} = \frac{V_2}{T_2} \V_2 = \frac{V_1 \times T_2}{T_1} = \frac{5 \times 500}{300} \taxminan 8.33\, \text{liters} degan ma'noni anglatadi.
\]
Muammo 7:
Monatomik gaz 2 atm da 6 litrdan 3 litrgacha siqilganda uning bajargan ishini toping.
Qarori:
Izobarik jarayonda bajarilgan ish uchun tenglamadan foydalanish:
\[
W = P \Delta V = 2 \times (3 – 6) = -6\, \text{atm}\cdot\text{liters}
\]
Muammo 8:
Ikki atomli gaz 300 K haroratda 3 litrdan 6 litrgacha izobar kengayishga uchraydi. Entropiyaning o'zgarishini toping.
Qarori:
Ikki atomli gaz uchun izobarik jarayonda entropiyaning o'zgarishi tenglamasidan foydalanib, bu yerda \( C_p = \frac{7}{2}R \):
\[
\Delta S = nC_p\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1} = 2 \times \frac{7}{2} \times 8.314 \times \ln\frac{300 \times \frac{6}{3}}{300} + 2 \times 8.314 \times \ln\frac{6}{3} \taxminan 34.76\, \text{J/K}
\]
Muammo 9:
500 K haroratda 10 litrdan 5 litrgacha bo'lgan 5 mol uch atomli gazning izobarik siqilishi paytida issiqlik uzatishni hisoblang.
Qarori:
Izobarik jarayonda issiqlik uzatish tenglamasidan foydalanish, bu yerda \(C_p = \frac{f}{2}R \) va \(f = 6 \):
\[
Q = 5 marta \frac{6}{2} \times 8.314 \times \left(500 \times \frac{5}{10} – 500\right) = -12473\, \text{J}
\]
Bu muammolar izobarik jarayon bilan bog'liq turli tushunchalar haqida keng qamrovli ma'lumot berishi kerak.
Izotermik jarayon (doimiy harorat) bilan bog'liq 9 ta muammo va yechimlar:
Muammo 1:
Masala: 300 K da 2 mol ideal gazning izotermik ravishda 1 litrdan 2 litrgacha kengayishi bilan bajarilgan ishni hisoblang.
Qarori:
Ideal gaz uchun izotermik jarayonda bajarilgan ish uchun tenglamadan foydalanib:
\[ W = nRT\ln\frac{V_2}{V_1} \]
\[
W = 2 \times 8.314 \times 300 \times \ln\frac{2}{1} \approx 3454\, \text{J}
\]
Muammo 2:
Masala: 400 K haroratda 3 mol gaz izotermik ravishda 4 litrdan 2 litrgacha siqilganda o'tkaziladigan issiqlik miqdorini toping.
Qarori:
Izotermik jarayonda issiqlik uzatish tenglamasidan foydalanish: \( Q = W \):
\[
Q = 3 \times 8.314 \times 400 \times \ln\frac{2}{4} \approx -3462\, \text{J}
\]
Muammo 3:
300 K haroratda 4 mol gaz 1 litrdan 5 litrgacha izotermik kengayishga uchraganda entropiyaning o'zgarishini aniqlang.
Qarori:
Izotermik jarayonda entropiyaning o'zgarishi tenglamasidan foydalanib:
\[
\Delta S = nR\ln\frac{V_2}{V_1} = 4 \times 8.314 \times \ln\frac{5}{1} \taxminan 46.15\, \text{J/K}
\]
Muammo 4:
200 K haroratda 1 mol ideal gaz 6 litrdan 3 litrgacha izotermik ravishda siqilganda bajarilgan ishni hisoblang.
Qarori:
Izotermik jarayonda bajarilgan ish uchun tenglamadan foydalanish:
\[
W = 1 \times 8.314 \times 200 \times \ln\frac{3}{6} \approx -575\, \text{J}
\]
Muammo 5:
Doimiy haroratda 2 litrdan 6 litrgacha izotermik kengayish paytida 3 mol gazning ichki energiyasining o'zgarishini aniqlang.
Qarori:
Ideal gazning izotermik jarayoni uchun ichki energiyaning o'zgarishi nolga teng:
\[
\Delta U = 0\, \text{J}
\]
Muammo 6:
250 K haroratda 5 mol gaz izotermik ravishda 3 litrdan 6 litrgacha kengayganda issiqlik uzatishni hisoblang.
Qarori:
Izotermik jarayonda issiqlik uzatish tenglamasidan foydalanish:
\[
Q = 5 \times 8.314 \times 250 \times \ln\frac{6}{3} \approx 2879\, \text{J}
\]
Muammo 7:
150 K haroratda 4 litrdan 2 litrgacha izotermik siqilish paytida 2 mol gaz qanday ish bajaradi?
Qarori:
Izotermik jarayonda bajarilgan ish uchun tenglamadan foydalanish:
\[
W = 2 \times 8.314 \times 150 \times \ln\frac{2}{4} \approx -1151\, \text{J}
\]
Muammo 8:
500 K haroratda 3 mol gaz 5 litrdan 1 litrgacha izotermik ravishda siqilganda entropiyaning o'zgarishini toping.
Qarori:
Izotermik jarayonda entropiyaning o'zgarishi tenglamasidan foydalanib:
\[
\Delta S = 3 \times 8.314 \times \ln\frac{1}{5} \taxminan -34.77\, \text{J/K}
\]
Muammo 9:
100 K da 4 mol ideal gazning izotermik ravishda 2 litrdan 8 litrgacha kengayishi bilan bajarilgan ishni aniqlang.
Qarori:
Izotermik jarayonda bajarilgan ish uchun tenglamadan foydalanish:
\[
W = 4 \times 8.314 \times 100 \times \ln\frac{8}{2} \approx 2304\, \text{J}
\]
Bu muammolar izotermik jarayon bilan bog'liq turli tushunchalarni, masalan, bajarilgan ish, issiqlik uzatish va entropiyaning o'zgarishini qamrab oladi.
Izoxorik jarayon (doimiy hajm) bilan bog'liq 9 ta muammo va yechimlar:
Muammo 1:
Doimiy hajmda harorat 200 K dan 400 K gacha ko'tarilganda, 3 mol monoatomli ideal gazning ichki energiyasining o'zgarishini hisoblang.
Qarori:
Doimiy hajmdagi ichki energiyaning o'zgarishi uchun tenglamadan foydalanish: \( \Delta U = nC_v\Delta T \), bu yerda \( C_v = \frac{3}{2}R \):
\[
\Delta U = 3 \times \frac{3}{2} \times 8.314 \times (400 – 200) \taxminan 3741\, \text{J}
\]
Muammo 2:
Doimiy hajmda harorat 300 K dan 200 K gacha tushirilganda 2 mol diatomik gaz uchun issiqlik uzatishni aniqlang.
Qarori:
Doimiy hajmda issiqlik uzatish tenglamasidan foydalanish: \( Q = \Delta U = nC_v\Delta T \), bu yerda \( C_v = \frac{5}{2}R \):
\[
Q = 2 marta \frac{5}{2} marta 8.314 marta (200 – 300) \taxminan -4157\, \text{J}
\]
Muammo 3:
Doimiy hajmda harorat 100 K dan 300 K gacha ko'tarilganda 4 mol uch atomli gaz uchun entropiyaning o'zgarishini hisoblang.
Qarori:
Doimiy hajmdagi entropiyaning o'zgarishi uchun tenglamadan foydalanib, bu yerda \( C_v = \frac{f}{2}R \) va \( f = 6 \):
\[
\Delta S = nC_v\ln\frac{T_2}{T_1} = 4 \times \frac{6}{2} \times 8.314 \times \ln\frac{300}{100} \taxminan 115.36\, \text{J/K}
\]
Muammo 4:
Izoxorik jarayon davomida gazning bajaradigan ishi nima?
Qarori:
Izoxorik jarayon davomida hajm doimiy bo'lgani uchun hech qanday ish bajarilmaydi:
\[
W = 0\, \text{J}
\]
Muammo 5:
5 mol monoatomli ideal gaz o'zgarmas hajmda 500 K dan 300 K gacha sovutilganda uning issiqlik o'tkazuvchanligini aniqlang.
Qarori:
Doimiy hajmdagi issiqlik uzatish tenglamasidan foydalanib, bu yerda \( C_v = \frac{3}{2}R \):
\[
Q = 5 marta \frac{3}{2} marta 8.314 marta (300 – 500) \taxminan -6232\, \text{J}
\]
Muammo 6:
Doimiy hajmda harorat 150 K dan 250 K gacha ko'tarilganda 1 mol diatomik gaz uchun ichki energiyaning o'zgarishini hisoblang.
Qarori:
Doimiy hajmdagi ichki energiyaning o'zgarishi uchun tenglamadan foydalanish, bu yerda \( C_v = \frac{5}{2}R \):
\[
\Delta U = 1 \times \frac{5}{2} \times 8.314 \times (250 – 150) \taxminan 2079\, \text{J}
\]
Muammo 7:
Doimiy hajmda harorat 600 K dan 300 K gacha pasayganda, 3 mol monoatomli gaz uchun entropiyaning o'zgarishini aniqlang.
Qarori:
Doimiy hajmdagi entropiyaning o'zgarishi uchun tenglamadan foydalanib, bu yerda \( C_v = \frac{3}{2}R \):
\[
\Delta S = 3 \times \frac{3}{2} \times 8.314 \times \ln\frac{300}{600} \taxminan -34.59\, \text{J/K}
\]
Muammo 8:
Doimiy hajmda harorat 200 K dan 400 K gacha ko'tarilganda 2 mol uch atomli gazning issiqlik o'tkazuvchanligi qanday?
Qarori:
Doimiy hajmdagi issiqlik uzatish tenglamasidan foydalanib, bu yerda \( C_v = \frac{6}{2}R \):
\[
Q = 2 marta \frac{6}{2} marta 8.314 marta (400 – 200) \taxminan 4986\, \text{J}
\]
Muammo 9:
4 mol monoatomli ideal gaz o'zgarmas hajmda 250 K dan 350 K gacha qizdirilganda uning ichki energiyasining o'zgarishini hisoblang.
Qarori:
Doimiy hajmdagi ichki energiyaning o'zgarishi uchun tenglamadan foydalanish, bu yerda \( C_v = \frac{3}{2}R \):
\[
\Delta U = 4 \times \frac{3}{2} \times 8.314 \times (350 – 250) \taxminan 4986\, \text{J}
\]
Bu muammolar izoxorik jarayonning turli jihatlarini, masalan, issiqlik uzatish, ichki energiyaning o'zgarishi va entropiyani turli xil gazlar uchun qamrab oladi.
