00:00Numerical number 5 of unit 15, a balloon contains 0.04 cubic meters of air at a pressure of 120
00:18kilopascal. Calculate the pressure required to reduce its volume to 0.025 meter cube at constant
00:29temperature. So, first of all, we will create data. Volume V1 is obtained as 0.04 meter cube
00:50or cubic meter. Pressure P1 is obtained as 120 kilopascal. We have to find pressure P2 and
01:12volume V2 is obtained as 0.025 meter cube. Here temperature is constant. So, T1 is equal
01:31to T2 is equal to T means constant. For calculation, we will use the formula as P1 V1 upon T1 is equal
01:55to P2 V2 upon T2. So, here temperature is constant. So, this formula will become as P1 V1 upon T is equal
02:10to P2 V2 upon T. So, T will be get cancelled with T. So, P1 V1 is equal to P2 V2. We have to find P2. So,
02:30we will rearrange this equation. So, after rearranging this equation will be P2 will be equal to P1 V1 upon V2.
02:42P1 120 kilopascal is equal to P1. So, P1 is equal to P1. So, P1 is equal to P1. So, P1 is equal to P1
02:49V1
03:01Cantonese. So, P1 is equal to P1 for P1 V2
03:08V1. V1 is 0.04 cubic meters divided by V2. V2 we obtained is 0.025. So, pressure P2 will be obtained is
03:26120 into 0.04 it will be 4.8 into 10 is power 3 divided by 0.025. So, 4.8 divided by 0.025 it will be
03:49192 into 10 is power 3 Pascal or we can also write this as 192 10 is power 3 Pascal is equal
04:12to kilo Pascal. So, it will be 192 kilo Pascal or further we can also write this as
04:25P2 P2 P2 P2 P2 will be 1.92 into 10 is power 5 Pascal. So, this is our final answer.
04:53So, this is our final answer.
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