Question

A gas contained in a box of 0.1 m^3 at atmospheric pressure is connected to another vessel of
0.09 m^3. Consequent change in pressure is X mm of Hg. Then X in metre is -

(1) 0.4 (2) 0.5 (3) 0.36 (4)0.3​

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Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{3.\:0.36\:m}}}$

$\rule{200}{4}$

$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

Given :-

Initial Volume = 0.1 m³

Final Volume = 0.1 + 0.09 = 0.19 m³

Now we know that

Pi. Vi = Pf. Vf

Where i & f denotes initial and final

Initial is atmospheric pressure that is 760 mm

$\bf\bold{P = \frac{760\times 0.1} {0.19} }$

Here P is final pressure

$\bf\bold{\Rightarrow \:P = 400\:mm}$

Therefore change in Pressure

$\bf\bold{\triangle \:P = 760 - 400}$

$\bf\bold{\triangle \:P = 360 \:mm}$

$\bf\bold{\triangle \:P = 0.36 \:m}$

Hence, Change in Pressure is 0.36 m

Answer 2

Answer:

Hii

Explanation:

shrujal.

How are you.