The pressure of a gas kept in an isothermal container is 200 KPa. If half of the gas is removed from it, the pressure will be :

800 K Pa

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400 K Pa

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200 K Pa

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100 K Pa

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Solution

The correct option is D 100 K Pa
Let the number of mole initially be n , and final is n/2

we know $P V = n r T$ ,T remains constant and volume of container if fix
So $200 \times V = n R T$ ...(1)

let the final pressure be P
$P V = n R T / 2$ ...(2)
From above equations , we get $P = 100 K P a$

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The pressure of a gas kept in an isothermal container is 200 KPa. If half of the gas is removed from it, the pressure will be :

The correct option is D 100 K PaLet the number of mole initially be n , and final is n/2we know PV=nrT ,T remains constant and volume of container if fixSo 200×V=nRT...(1)let the final pressure be PPV=nRT/2...(2)From above equations , we get P=100 KPa

The correct option is D 100 K Pa
Let the number of mole initially be n , and final is n/2

we know $P V = n r T$ ,T remains constant and volume of container if fix
So $200 \times V = n R T$ ...(1)

let the final pressure be P
$P V = n R T / 2$ ...(2)
From above equations , we get $P = 100 K P a$