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Pertanyaan

Sejumlah gas diatomik pada suhu 120°C dan tekanan 8 × 1 0 4 N / m 2 mempunyai volume 4 L Gas ini mengalami proses isobarik hingga volumenya 4,8 Llalu proses isokhorik hingga tekananya 6 × 1 0 4 N / m 2 . Berapa perubahan energi dalamnya?

Sejumlah gas diatomik pada suhu 120°C dan tekanan   mempunyai volume 4 L Gas ini mengalami proses isobarik hingga volumenya 4,8 L lalu proses isokhorik hingga tekananya . Berapa perubahan energi dalamnya?

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Y. Maghfirah

Master Teacher

Jawaban terverifikasi

Jawaban

perubahan energi dalamnya -80J.

perubahan energi dalamnya -80 J.

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Pembahasan

Diketahui : Ditanya : ? Jawab : dan dihitung dengan rumus Boyle-Gay Lussac. Proses AB : Isobarik (Tekanan tetap) Proses BC: Isokhorik (Volume tetap) Untuk menghitung n kita bisa gunakan persamaan gas umum Untuk gas diatomik Proses AB : Isobarik Proses BC : Isokhorik Perubahan energi dalam Jadi, perubahan energi dalamnya -80J.

Diketahui :

  begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript A end cell equals cell 4 space straight L equals 4 cross times 10 to the power of negative 3 end exponent space straight m cubed end cell row cell V subscript B end cell equals cell 4 comma 8 space straight L equals 4 comma 8 cross times 10 to the power of negative 3 end exponent space straight m cubed end cell row cell V subscript C end cell equals cell V subscript B equals 4 comma 8 space straight L equals 4 comma 8 cross times 10 to the power of negative 3 end exponent space straight m cubed end cell row cell P subscript A end cell equals cell 8 cross times 10 to the power of 4 space straight N divided by straight m to the power of 4 end cell row cell P subscript B end cell equals cell straight P subscript straight A equals 8 cross times 10 to the power of 4 space straight N divided by straight m to the power of 4 end cell row cell P subscript C end cell equals cell 6 cross times 10 to the power of 4 space straight N divided by straight m squared end cell row cell T subscript A end cell equals cell 120 plus 273 equals 393 space straight K end cell end table end style 

Ditanya : begin mathsize 14px style increment U end style ?

Jawab :

begin mathsize 14px style T subscript C end style dan begin mathsize 14px style T subscript B end style dihitung dengan rumus Boyle-Gay Lussac.

Proses AB : Isobarik (Tekanan tetap)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator P subscript A V subscript A over denominator T subscript A end fraction end cell equals cell fraction numerator P subscript B V subscript B over denominator T subscript B end fraction end cell row cell V subscript A over T subscript A end cell equals cell V subscript B over T subscript B end cell row cell T subscript B end cell equals cell V subscript B over V subscript A T subscript A end cell row cell T subscript B end cell equals cell fraction numerator 4 comma 8 cross times 10 to the power of negative 3 end exponent over denominator 4 cross times 10 to the power of negative 3 end exponent end fraction 393 end cell row cell T subscript B end cell equals cell 471 comma 6 space straight K end cell end table end style 

Proses BC: Isokhorik (Volume tetap)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator P subscript C V subscript C over denominator T subscript C end fraction end cell equals cell fraction numerator P subscript B V subscript B over denominator T subscript B end fraction end cell row cell P subscript C over T subscript C end cell equals cell P subscript B over T subscript B end cell row cell T subscript C end cell equals cell P subscript C over P subscript B T subscript B end cell row cell T subscript C end cell equals cell fraction numerator 6 cross times 10 to the power of 4 over denominator 8 cross times 10 to the power of 4 end fraction cross times 471 comma 6 end cell row cell T subscript C end cell equals cell 353 comma 7 space straight K end cell end table end style 

Untuk menghitung n kita bisa gunakan persamaan gas umum

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell P subscript A V subscript A end cell equals cell n R T subscript A end cell row n equals cell fraction numerator P subscript A V subscript A over denominator R T subscript A end fraction equals fraction numerator 8 cross times 10 to the power of 4.4 cross times 10 to the power of negative 3 end exponent over denominator 8 comma 31.393 end fraction equals 0 comma 098 space mol end cell end table end style 

Untuk gas diatomik  

begin mathsize 14px style C subscript V equals 5 over 2 R equals 5 over 2.8 comma 31 equals 20 comma 775 space straight J divided by mol. straight K end style 

Proses AB : Isobarik

begin mathsize 14px style increment U subscript A B end subscript equals n C subscript V left parenthesis T subscript B minus T subscript A right parenthesis increment U subscript A B end subscript equals 0 comma 098 cross times 20 comma 775 left parenthesis 471 comma 6 minus 393 right parenthesis increment U subscript A B end subscript equals 160 space straight J end style 

Proses BC: Isokhorik

begin mathsize 14px style increment U subscript A B end subscript equals n C subscript V left parenthesis T subscript C minus T subscript B right parenthesis increment U subscript A B end subscript equals 0 comma 098 cross times 20 comma 775 left parenthesis 353 comma 7 minus 471 comma 6 right parenthesis increment U subscript A B end subscript equals negative 240 space straight J end style  

Perubahan energi dalam

begin mathsize 14px style increment U equals increment U subscript A B end subscript plus increment U subscript B C end subscript increment U equals 160 minus 240 increment U equals negative 80 space straight J end style 

Jadi, perubahan energi dalamnya -80 J.

Latihan Bab

Hukum Ke - Nol dan Hukum I Termodinamika

Proses Keadaan Gas

Kapasitas Kalor

Hukum II Termodinamika

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