Question

N2 is found in a litre flask under 100 k Pa
pressure and O2 is found in another 3 litre
flask under 320 kPa pressure. If the two flasks
are connected, the resultant pressure is:
(1 310 kPa (2) 210 kPa
(3) 420 kPa (4) 265 kPa​

Answer 1

Answer:-

$\bold{P = 265Kpa}$

Expanation:-

Given:-

• $P_{O_2} = 320kpa$
• $P_{N_2} = 100kpa$
• $V_{O_2} = 3 L$
• $V_{N_2} = 1 L$

To Find:-

Pressure of the mixture

Solution:-

Let, resultant pressure be P

Now,

$\implies P_{O_2}V_{O_2}+P_{N_2}V_{N_2}= PV_{O_2}+P V_{N_2}$

$\implies( 320)(3)+(100)(1)= P(3)+P (1)$

$\implies 960+100= P(3 +1)$

$\implies 1060 = 4P$

$\implies \bold{P = 265Kpa}$_____option(4)

Answer 2

$\huge \sf \green{Hello \: mate}$

Given:-

$P_{O_2} = 320kpaP$

$P_{N_2} = 100kpaP$

$V_{O_2} = 3 LV$

$V_{N_2} = 1 LV$

To Find:-

Pressure of the mixture

Solution:-

Let, resultant pressure be P

Now,

$\implies P_{O_2}V_{O_2}+P_{N_2}V_{N_2}= PV_{O_2}+P V_{N_2}⟹P$

$\implies( 320)(3)+(100)(1)= P(3)+P (1)⟹(320)(3)+(100)(1)=P(3)+P(1)$

$\implies 960+100= P(3 +1)⟹960+100=P(3+1)$

$\implies 1060 = 4P⟹1060=4P$

$\implies \bold{P = 265Kpa}$

⟹P=265Kpa _____option(4)