A system undergoes an isobaric process where the pressure is kept constant at 150 kPa. If the volume increases from 0.05 m³ to 0.08 m³. Calculate the heat added to the system

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A system undergoes an isobaric process where the pressure is kept constant at 150 kPa. If the volume increases from 0.05 m³ to 0.08 m³. Calculate the heat added to the system Physics class 12 numericals  First law of thermodynamics numericals  12 class numerical solution  New Sindh Text Physics numericals

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00:00Numerical number 3 of unit 16 first law of thermodynamics is a system undergoes an isobaric

00:13process where the pressure is kept constant at 150 kilopascal. If the volume increases

00:21from 0.05 meter cube to 0.08 meter cube calculate heat added to the system. So,

00:33we will create data. P1 is obtained as 150 kilopascal. So,

00:56P1 will be equal to 150, means one lake 50,000 pascal.

01:11کیونکہ 1 kilopascal جو ہوتا ہے وہ 1000 pascal کے ایکیول ہوتا ہے.

01:18Or V1 is obtained as 0.05 meter cube. V2 is obtained as 0.08 meter cube. So, we have to find heat supplied

01:43means delta q is equal to watt. یہاں پر internal energy absent ہے.

01:53So, internal energy کا جو value absent ہے that is 1500 joules.

02:03So, for calculation

02:05we will use the formula is here the process is isobaric process. So, according to isobaric process,

02:20the equation of first law of thermodynamics will be delta q is equal to delta u plus p delta v

02:32minus v2 minus v1. So, this equation will become is delta q will be equal to delta u plus p into

02:50v2 minus v1. So, by putting the values delta q will be equal to internal energy is obtained as

02:581500 joules plus 1500 joules plus pressure is 1500 joules plus pressure is one lake 50,000

03:08into v2. v2 is 0.08 meter cube minus v1 is 0.05 meter cube.

03:17So, delta q will be equal to 1500 filling. So, delta q will be equal to 1500 joules plus

03:29into 0.08 minus 0.05 it will be 0.03 nogetugaала塊 etc.

03:36So, delta q will be equal to 1500 joules plus

03:42so delta q will be equal to 1500 joules plus

03:43when we will multiply one leg 50,000 with 0.03 then it will be 4500.

03:56So, delta Q will be equal to 1500 plus 4500 it will be 6000 joules. So, this is our required answer

04:09is in textbook the answer is given is 600 joules but the actual answer is 6000 joules.

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