Adeegsiga sharciga koowaad ee thermodynamics-ka qaar ka mid ah hababka thermodynamic-ka (Isobaric Isothermal Isochoric)

30 Adeegsiga sharciga koowaad ee thermodynamics-ka qaar ka mid ah hababka thermodynamic-ka (Isobaric Isothermal Isochoric)

1. Jaantuska hoose wuxuu muujinayaa wareegga thermodynamic- ka ee uu gaas la kulmo. Shaqada uu gaasku ka qabto habka ABCD waa…

La yaqaan:

Cadaadiska 1 (P ₁ ) = 2 x 10 ⁵ Pa

Cadaadiska 2 (P ₂ ) = 4 x 10 ⁵ Pa

Mugga 1 (V1 ₎ = 1 m ³

Mugga 2 (V2 ₎ = 3 m ³

La Doonayo: Shaqada uu gaaska ku sameeyo habka ABCD (W).

Xalka:

Shaqada uu sameeyo gaasku waxay la mid tahay aagga ABCD.

W = (P ₂ – P ₁ )(V ₂ – V ₁ )

W = (4 x 10 ⁵ – 2 x 10 ⁵ )(3 – 1)

W = (2 x 10 ⁵ )(2)

W = 4 x 10 ⁵ Joule

2. Shaqada uu gaaska ka qabto habka ABC waa…

La yaqaan:

Cadaadiska 1 (P ₁ ) = 3 x 10 ⁵ Pa

Cadaadiska 2 (P ₂ ) = 6 x 10 ⁵ Pa

Mugga 1 (V1 ₎ = 20 cm ³ = 20 x 10 ^-6 m ³

Mugga 2 (V2 ₎ = 60 cm ³ = 60 x 10 ^-6 m ³

La Doonayo : Shaqada uu gaaska ka qabtay habka ABC

Xalka:

Shaqada waxaa qabta gaaska = aagga ABC.

W = ½ (P ₂ – P ₁ )(V ₂ – V ₁ )

W = 1/2 (6 x 10 ⁵ – 3 x 10 ⁵ )(60 x 10 ^-6 – 20 x 10 ^-6 )

W = 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 Joule

3. Jaantuska PV ee gaaska ku jira weel xiran oo lagu muujiyay sawirka hoose. Shaqada uu gaaska sameeyay waxaa lagu muujiyay habka loo marayo.

Solution

Habka AB iyo habka DC waa hababka isobaric (cadaadiska waa la ilaaliyaa si joogto ah) . Habka AD iyo habka BC waa hababka choric ( mugga isma beddelo ).

Shaqada waxaa lagu sameeyay gaas markii gaasku ballaarto (habka DC).

Waa kuwan 9 dhibaato iyo xalal la xiriira Habka Isobaric (Cadaadiska Joogtada ah):

Dhibaatada 1:
Gaas 3.0 m³ ah oo cadaadis joogto ah leh ayaa lagu cadaadiyaa ilaa 2.0 m³. Haddii heerkulka bilowga ahi yahay 300 K, waa maxay heerkulka ugu dambeeya?

Solution:
Annagoo adeegsanayna isla'egta habka isobaric: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), waxaan xallin karnaa heerkulka ugu dambeeya:

\[
\frac{3.0}{300} = \frac{2.0}{T_2} \tilmaamaya T_2 = \frac{2.0 \times 300}{3.0} = 200\, \text{K}
\]

Dhibaatada 2:
Xisaabi shaqada ay qabteen 2 moles oo gaas ah oo ku fidaya isobar ahaan laga bilaabo 1 litir ilaa 3 litir 2 atm.

Solution:
Isticmaalka isla'egta shaqada lagu sameeyo habka isobaric: \( W = P \Delta V \):

\[
W = 2\, \text{atm} \times (3 – 1)\, \text{liters} = 4\, \text{atm}\cdot\text{liters}
\]

Dhibaatada 3:
Waa maxay isbeddelka tamarta gudaha marka 4 moles oo gaas diatomic ah oo ku habboon lagu kululeeyo cadaadis joogto ah laga bilaabo 200 K ilaa 300 K?

Solution:
Isticmaalka isleegga isbeddelka tamarta gudaha ee cadaadiska joogtada ah ee gaaska diatomic: \( \Delta U = nC_pdT \), halkaas oo \( C_p = \frac{7}{2}R \):

\[
\Delta U = 4 \jeer \frac{7}{2} \jeer 8.314 \jeer (300 - 200) \qiyaastii 6990\, \text{J}
\]

Dhibaatada 4:
Xisaabi kulaylka lagu daray inta lagu jiro habka isobaric marka 2 moles oo gaas monatomic ah ay ka fidaan 5 litir ilaa 10 litir 400 K.

Solution:
Isticmaalka isle'egta wareejinta kulaylka ee habka isobaric: \( Q = nC_pdT \), halkaas oo \( C_p = \frac{5}{2}R \), iyo \( \frac{V_2}{T_2} = \frac{V_1}{T_1} \):

\[
Q = 2 \jeer \frac{5}{2} \jeer 8.314 \jeer \left(400 \jeer \frac{10}{5} – 400\right) = 3316\, \text{J}
\]

Dhibaatada 5:
Go'aami isbeddelka ku yimaada enthalpy ee 3 moles oo gaas ah oo saddex-geesood ah oo laga ballaariyay 1 litir ilaa 2 litir heerkulkiisu yahay 200 K.

Solution:
Isticmaalka qaacidada isbeddelka enthalpy ee habka isobaric: \( \Delta H = nC_pdT \), halkaas oo \( C_p = \frac{f}{2}R \) iyo \( f = 6 \):

\[
\Delta H = 3 \jeer \frac{6}{2} \jeer 8.314 \jeer \left(200 \jeer \frac{2}{1} – 200\right) = 4989\, \text{J}
\]

Dhibaatada 6:
Xisaabi mugga ugu dambeeya ee 2 moles oo gaas ah 300 K iyo 5 litir marka hore marka la kululeeyo 500 K cadaadis joogto ah.

Solution:
Isticmaalka isla'egta habka isobaric:

\[
\frac{V_1}{T_1} = \frac{V_2}{T_2} \ikhtiraacyada V_2 = \frac{V_1 \times T_2}{T_1} = \frac{5 \times 500}{300} \qiyaastii 8.33\, \text{litir}
\]

Dhibaatada 7:
Soo hel shaqada uu qabtay gaaska monatomic marka la cadaadiyo laga bilaabo 6 litir ilaa 3 litir 2 atm.

Solution:
Isticmaalka isla'egta shaqada lagu sameeyo habka isobaric:

\[
W = P \Delta V = 2 \ jeer (3 - 6) = -6\, \text{atm}\cdot\text{litir}
\]

Dhibaatada 8:
Gaaska diatomic-ga wuxuu maraa ballaarin isobaric ah laga bilaabo 3 litir ilaa 6 litir 300 K. Soo hel isbeddelka entropy-ga.

Solution:
Isticmaalka isleegga isbeddelka entropy ee habka isobaric ee gaaska diatomic, halkaas oo \(C_p = \frac{7}{2}R \):

\[
\Delta S = nC_p\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1} = 2 \times \frac{7}{2} \times 8.314 \times \ln\frac{300 \times \frac{6}{3}}{300} + 2 \times 8.314 \times \ln\frac{6}{3} \ku dhawaad ​​34.76\, \text{J/K}
\]

Dhibaatada 9:
Xisaabi wareejinta kulaylka inta lagu jiro cadaadiska isobaric ee 5 moles oo gaas triatomic ah laga bilaabo 10 litir ilaa 5 litir 500 K.

Solution:
Isticmaalka isle'egta wareejinta kulaylka ee habka isobaric, halkaas oo \(C_p = \frac{f}{2}R \) iyo \(f = 6 \):

\[
Q = 5 \jeer \frac{6}{2} \jeer 8.314 \jeer \left(500 \jeer \frac{5}{10} – 500\right) = -12473\, \text{J}
\]

Dhibaatooyinkani waa inay bixiyaan dulmar guud oo ku saabsan fikradaha kala duwan ee la xiriira habka isobaric.

Waa kuwan 9 dhibaato iyo xalal la xiriira Habka Isothermal-ka (Heerkulka Joogtada ah):

 Dhibaatada 1:
Dhibaato: Xisaabi shaqada ay qabteen 2 moles oo gaas ku habboon ah oo ku fidaya meel fog laga bilaabo 1 litir ilaa 2 litir oo ah 300 K.

Solution:
Isticmaalka isla'egta shaqada lagu qabtay habka isothermal-ka ee gaaska ku habboon:
\[ W = nRT\ln\frac{V_2}{V_1} \]

\[
W = 2 \jeer 8.314 \jeer 300 \jeer \ln\frac{2}{1} \qiyaastii 3454\, \text{J}
\]

Dhibaatada 2:
Dhibaato: Soo hel kulaylka la wareejiyay marka 3 moles oo gaas ah lagu cadaadiyo meel ka baxsan heerkulka laga bilaabo 4 litir ilaa 2 litir heerkul ah 400 K.

Solution:
Isticmaalka isle'egta wareejinta kulaylka ee habka isothermal-ka: \( Q = W \):

\[
Q = 3 \jeer 8.314 \jeer 400 \jeer \ln\frac{2}{4} \approx -3462\, \text{J}
\]

Dhibaatada 3:
Go'aami isbeddelka entropy-ga marka 4 moles oo gaas ah ay maraan ballaarinta isothermal laga bilaabo 1 litir ilaa 5 litir heerkul ah 300 K.

Solution:
Isticmaalka isle'egta isbeddelka entropy ee habka isothermal-ka:

\[
\Delta S = nR\ln\frac{V_2}{V_1} = 4 \jeer 8.314 \jeer \ln\frac{5}{1} \qiyaastii 46.15\, \text{J/K}
\]

Dhibaatada 4:
Xisaabi shaqada la qabtay marka 1 mole oo gaas ah oo ku habboon la cadaadiyo si gooni gooni ah laga bilaabo 6 litir ilaa 3 litir oo ah 200 K.

Solution:
Isticmaalka isla'egta shaqada lagu qabtay habka isothermal-ka:

\[
W = 1 \jeer 8.314 \jeer 200 \jeer \ln\frac{3}{6} \qiyaastii -575\, \qoraal{J}
\]

Dhibaatada 5:
Go'aami isbeddelka tamarta gudaha ee 3 moles oo gaas ah inta lagu jiro ballaarinta isothermal laga bilaabo 2 litir ilaa 6 litir heerkul joogto ah.

Solution:
Habka isothermal-ka ee gaaska ku habboon, isbeddelka tamarta gudaha waa eber:

\[
\Delta U = 0\, \qoraal{J}
\]

Dhibaatada 6:
Xisaabi wareejinta kulaylka marka 5 moles oo gaas ah lagu ballaariyo meel ka baxsan heerkulka laga bilaabo 3 litir ilaa 6 litir oo ah 250 K.

Solution:
Isticmaalka isle'egta wareejinta kulaylka ee habka isothermal-ka:

\[
Q = 5 \jeer 8.314 \jeer 250 \jeer \ln\frac{6}{3} \qiyaastii 2879\, \text{J}
\]

Dhibaatada 7:
Waa maxay shaqada ay qabtaan 2 moles oo gaas ah inta lagu jiro cadaadiska isothermal-ka laga bilaabo 4 litir ilaa 2 litir 150 K?

Solution:
Isticmaalka isla'egta shaqada lagu qabtay habka isothermal-ka:

\[
W = 2 \jeer 8.314 \jeer 150 \jeer \ln\frac{2}{4} \qiyaastii -1151\, \qoraal{J}
\]

Dhibaatada 8:
Soo hel isbeddelka entropy marka 3 moles oo gaas ah la cadaadiyo si gooni gooni ah laga bilaabo 5 litir ilaa 1 litir heerkul ah 500 K.

Solution:
Isticmaalka isle'egta isbeddelka entropy ee habka isothermal-ka:

\[
\Delta S = 3 \jeer 8.314 \jeer \ln\frac{1}{5} \approx -34.77\, \text{J/K}
\]

Dhibaatada 9:
Go'aami shaqada ay qabteen 4 moles oo gaas ku habboon oo ku fidaya meel fog laga bilaabo 2 litir ilaa 8 litir 100 K.

Solution:
Isticmaalka isla'egta shaqada lagu qabtay habka isothermal-ka:

\[
W = 4 \jeer 8.314 \jeer 100 \jeer \ln\frac{8}{2} \qiyaastii 2304\, \text{J}
\]

Dhibaatooyinkani waxay daboolayaan fikrado kala duwan oo la xiriira habka isothermal-ka, sida shaqada la qabtay, wareejinta kulaylka, iyo isbeddelka entropy-ga.

Waa kuwan 9 dhibaato iyo xalal la xiriira Habka Isochoric (Mugga Joogtada ah):

Dhibaatada 1:
Xisaabi isbeddelka tamarta gudaha ee 3 moles oo gaas ah oo ku habboon monatomic marka heerkulka la kordhiyo laga bilaabo 200 K ilaa 400 K mugga joogtada ah.

Solution:
Isticmaalka isleegga isbeddelka tamarta gudaha mugga joogtada ah: \( \Delta U = nC_v\Delta T \), halkaas oo \( C_v = \frac{3}{2}R \):

\[
\Delta U = 3 \jeer \frac{3}{2} \jeer 8.314 \jeer (400 - 200) \qiyaastii 3741\, \text{J}
\]

Dhibaatada 2:
Go'aami wareejinta kulaylka ee 2 moles oo gaas diatomic ah marka heerkulku hoos u dhaco laga bilaabo 300 K ilaa 200 K marka mugga joogtada ah la gaaro.

Solution:
Isticmaalka isle'egta wareejinta kulaylka mugga joogtada ah: \( Q = \Delta U = nC_v\Delta T \), halkaas oo \( C_v = \frac{5}{2}R \):

\[
Q = 2 \jeer \frac{5}{2} \jeer 8.314 \jeer (200 - 300) \qiyaastii -4157\, \text{J}
\]

Dhibaatada 3:
Xisaabi isbeddelka entropy ee 4 moles oo gaas ah oo saddex-geesood ah marka heerkulku ka kordho 100 K ilaa 300 K mugga joogtada ah.

Solution:
Isticmaalka isleegga isbeddelka entropy-ga mugga joogtada ah, halkaas oo \(C_v = \frac{f}{2}R \) iyo \( f = 6 \):

\[
\Delta S = nC_v\ln\frac{T_2}{T_1} = 4 \times \frac{6}{2} \times 8.314 \times \ln\frac{300}{100} \ku dhawaad ​​115.36\, \text{J/K}
\]

Dhibaatada 4:
Waa maxay shaqada uu qabto gaaska inta lagu jiro habka isochoric?

Solution:
Maadaama mugga uu yahay mid joogto ah inta lagu jiro habka isochoric, ma jiro wax shaqo ah oo la qabtay:

\[
W = 0\, \qoraal{J}
\]

Dhibaatada 5:
Go'aami wareejinta kulaylka ee 5 moles oo gaas ah oo ku habboon monatomic marka la qaboojiyo laga bilaabo 500 K ilaa 300 K mugga joogtada ah.

Solution:
Iyadoo la adeegsanayo isle'egta wareejinta kulaylka mugga joogtada ah, halkaas oo \(C_v = \frac{3}{2}R \):

\[
Q = 5 \jeer \frac{3}{2} \jeer 8.314 \jeer (300 - 500) \qiyaastii -6232\, \text{J}
\]

Dhibaatada 6:
Xisaabi isbeddelka tamarta gudaha ee 1 mole oo gaas diatomic ah marka heerkulka la kordhiyo laga bilaabo 150 K ilaa 250 K mugga joogtada ah.

Solution:
Isticmaalka isleegga isbeddelka tamarta gudaha mugga joogtada ah, halkaas oo \(C_v = \frac{5}{2}R \):

\[
\Delta U = 1 \jeer \frac{5}{2} \jeer 8.314 \jeer (250 - 150) \qiyaastii 2079\, \text{J}
\]

Dhibaatada 7:
Go'aami isbeddelka entropy ee 3 moles oo gaas monatomic ah marka heerkulku hoos uga dhaco 600 K ilaa 300 K mugga joogtada ah.

Solution:
Isticmaalka isla'egta isbeddelka entropy-ga mugga joogtada ah, halkaas oo \( C_v = \frac{3}{2}R \):

\[
\Delta S = 3 \jeer \frac{3}{2} \jeer 8.314 \jeer \ln\frac{300}{600} \approx -34.59\, \text{J/K}
\]

Dhibaatada 8:
Waa maxay wareejinta kulaylka ee 2 moles oo gaas triatomic ah marka heerkulku ka kordho 200 K ilaa 400 K mugga joogtada ah?

Solution:
Iyadoo la adeegsanayo isle'egta wareejinta kulaylka mugga joogtada ah, halkaas oo \(C_v = \frac{6}{2}R \):

\[
Q = 2 \jeer \frac{6}{2} \jeer 8.314 \jeer (400 - 200) \qiyaastii 4986\, \text{J}
\]

Dhibaatada 9:
Xisaabi isbeddelka tamarta gudaha ee 4 moles oo gaas ah oo ku habboon monatomic marka la kululeeyo laga bilaabo 250 K ilaa 350 K mugga joogtada ah.

Solution:
Isticmaalka isleegga isbeddelka tamarta gudaha mugga joogtada ah, halkaas oo \(C_v = \frac{3}{2}R \):

\[
\Delta U = 4 \jeer \frac{3}{2} \jeer 8.314 \jeer (350 - 250) \qiyaastii 4986\, \text{J}
\]

Dhibaatooyinkani waxay daboolayaan dhinacyo kala duwan oo ka mid ah habka isochoric, sida wareejinta kulaylka, isbeddellada tamarta gudaha, iyo entropy, noocyada kala duwan ee gaasaska.