Question

At constant temperature the pressure of 22.4dm^3 volume of an ideal gas was increased from 105kPa to 210kPa, New volume could be?

Answer 1

Answer:

11.2dm³.

Explanation:

p1*v1=p2*v2

105*22.4=210*v2

v2=11.2dm³.

Answer 2

Answer:

$V_{2} = 11.2$ dm^3.

Explanation:

Since we know that for an isotherm process (When temperature of the process remains constant) and for a unit mole ,the relation between two gases are given by,

$P_{1}V_{1} =P_{2}V_{2}$...............................(1)

Where P = Pressure of the gas , V = volume of the gas

1 and 2 represents the two gases.

we have $P_{1} = 105$ kPa ,$P_{2} = 210$ KPa, $V_{1} = 22.4$ dm^3.

Using equation (1) we get

$105\times 22.4 = 210\times V_{2}$

So, new volume

$V_{2} = 11.2$ dm^3.