An air storage tank whose volume is 112 litres contain 3kg of air at a pressure of 18 atmosphere. How much air would have to be forced into the tank to increase the pressure to 21 atmospheres, assuming no change in temperature

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00:00The numerical number 4 of unit 15 is an air storage tank whose volume is 112 liters, contain

00:173 kilograms of air at a pressure of 18 atmospheres. How much air would have to be forced into

00:25the tank to increase the pressure to 21 atmospheres, assuming no change in temperature. So, first

00:36of all, we make data. Volume which we have obtained is V which is 112 liters.

00:52Pressure P1, we have obtained 18 atmospheres. Pressure P2, we have obtained 21 atm, means atmospheres.

01:13Initial mass of air is M1 which is 3 kilogram. So, we have to find final mass of air means M2 is equal to

01:35is equal to Watt. And here, temperature T1 is equal to T2 is equal to T means temperature is constant.

01:47So, for calculation, we will use formula P1 V1 upon M1 is equal to P2 V2 upon M2 is from given condition,

02:10volume is constant. So, V1 is equal to V2 is equal to V. So, P1 V upon M1 is equal to P2 V upon M2.

02:26Here, we have to find M2. So, V1 is equal to M2. Here, we have to find M2. So,

02:42we have to find this equation. So, we have to find this equation. So, we have to find this equation.

02:46M2 will be equal to M2 will be equal to P2 upon P1 into M1.

02:54Now, putting the given values, M2 will be equal to P2 is obtained as 21 into M1.

03:10M1 is 3 kg divided by P1. P1 is 18 atmospheres. So, M2 will be obtained as

03:403.5 kg. The amount of air to be added is the difference between the final mass and the initial mass.

04:00So, air to be added will be equal to M2 minus M1. So, air to be added will be as M2 is 3.5 kg minus M1. M1 is 3 kg.

04:24So, air to be added will be 3.5 minus 3. So, it will be 0.5 kg. So, this is our final answer.

04:39So, where do we go?

04:42I can not complain.

04:45So, are you fears