Foydalanuvchi:Yulduz Nazarova O'zMU/qumloq

Adiabatik jarayonda bajarilgan ish

1 mol gazning V1 dan V2 gacha kengayganda bajarilgan ishning hisoblaymiz.

$dA=PdV$ ( 1)

$A=\int _{V_{1}}^{V_{2}}PdV$

$P\sim V$ adiabata bo'yicha bog'langan, ya'ni $PV^{\gamma }=const$ (2)

$PV^{\gamma }=P_{1}V_{1}^{\gamma }$ (3)

$P={P_{1}V_{1}^{\gamma } \over V^{\gamma }}$

bu yerda V1 – boshlang'ich hajm.

$A=\int _{V_{1}}^{V_{2}}P_{1}V_{1}^{\gamma }{dV \over V^{\gamma }}=$ $P_{1}1V_{1}^{\gamma }\int _{V_{2}}^{V_{1}}{dV \over dV^{\gamma }}$

$A=P_{1}V_{1}^{\gamma }{1 \over -\gamma +1}{\biggl (}-{1 \over {V_{1}^{\gamma } \over V_{1}}}+{1 \over {V_{2}^{\gamma } \over V_{2}}}{\biggr )}$ $={P_{1}V_{1}^{\gamma } \over -\gamma +1}{1 \over {V_{1}^{\gamma } \over V_{1}}}[1-{\bigl (}{V_{1} \over V_{2}}{\bigr )}$ ^ ${\bigl (}\gamma -1{\bigr )}]$

$P_{1}V_{1}=RT_{1}$ ekanligini hisobga olsak:

$A={P_{1}V_{1} \over \gamma -1}{\biggl (}1-{\bigl (}{V_{1} \over V_{2}}{\bigr )}$ ^ ${\bigl (}\gamma -1{\bigr )}{\biggr )}$ $={RT_{1} \over \gamma -1}[1-{\bigl (}{V_{1} \over V_{2}}{\bigr )}$ ^ ${\bigl (}\gamma -1{\bigr )}]$

Agar ${M \over \mu }$ nol gaz uchun yozsak

$A={M \over \mu }{RT_{1} \over \gamma -1}[1-{\bigl (}{V_{1} \over V_{2}}{\bigr )}$ ^ ${\bigl (}\gamma -1{\bigr )}]$ (4)

Boshqa ko'rinishdan foydalansak, ${T_{1} \over T_{2}}={\bigl (}{V_{2} \over V_{1}}{\bigr )}$ ^ ${\bigl (}\gamma -1{\bigr )}$

Shuning uchun $A={MRT_{1} \over \mu {\bigl (}\gamma -1{\bigr )}}{\Bigl (}1-{T_{2} \over T_{1}}{\Bigr )}=$ ${MR \over \mu {\bigl (}\gamma -1{\bigr )}}{\bigl (}T_{1}-T_{2}{\bigr )}$ (5)