Question

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$<marquee>a vessel of 120 ml capacity contain a certain amount of gas at 35 degree Celsius and 1.2 bar pressure the gas is transferred to another vessel of volume 180 ml at 35 degree celsius what would be its pressure$​

Answer 1

Explanation:

Here , P1=1.2 bar, V1=120 ml , P2=?, V2=180 ml

temp of gas is at 35°c

e.g. temp is constant

we have apply Boyle's law

P1V1 = P2V2

Now,

1.2 × 120 = P2 × 180

=> 144 = P2 × 180

=> P2 × 180 = 144

=> P2 = 144/180 = 0.8

Answer 2

Answer:

Answer

The volume would be 120mL and it shows that it is constant.

Explanation:

Given parameters:

Intial volume of gas V₁ = 800mL

Initial temperature T₁ = -35°C to Kelvin; we use K = °C + 273 = -23 + 273 = 250K

Inital pressure P₁ = 300torr

Final temperature T₂ = 227°C to K = 227 + 273 = 500K

Final pressure P₂ = 600torr

Unknown parameter:

Final volume V₂ = ?

To solve this problem, we use the General gas law or the combined gas law where we assume that n=1, this gives:

\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }

Since the unknown is V₂, we make it the subject of the formula:

V₂ = \frac{P_{1} V_{1} T_{2} }{P_{2} T_{1} }

V₂ = \frac{300x800x500}{600x250}

V₂ = 800mL