Question

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A vessel is filled with a gas at a pressure of 76cm of mercury at a certain temperature.The mass of the gas is increased by 50% by introducing more gas in the vessel at the same temperature. The resultant pressure of the gas is?​

Answer 1

Answer:

114cm

Explanation:

We will use this famous equation, (personally, I love this equation,)

$PV=nRT$

Conditions:

1. constant temperature (given)
2. constant volume, the vessel is same
3. gas constant is a constant in itself,so,DUH!

Equation modification :  $P_{1} .n_{2} =P_{2} .n_{1}$ (1 and 2 represent initial and final situations respectively)

$n_{1} =n$ and $n_{2} =n+0.5n=1.5n$

Substitute and solve for $P_{2}$.

Comes out to be, 114cm!

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Kudos!

Answer 2

Answer:

$\huge\sf\purple{Answer}$

Explanation:

➪Given☟︎︎︎

A vessel is filled with a gas at a pressure of 76cm of mercury at a certain temperature.The mass of the gas is increased by 50% by introducing more gas in the vessel at the same temperature. The resultant pressure of the gas is?

➪To find☟︎︎︎

• ➪P2 = ?

• ➪T1=T2=T

➪Solution☟︎︎︎

• ➪Let initial mass of the gas be m1

• ➪Increase in mass Am/m1 = 0.5

• ➪Thus final mass m2 = m1 + 0.5m1 = 1.5m1

• ➪Let M be the molecular mass of the gas

• ➪Thus, number of moles in the original

• ➪mass of gas n1 = m1/M

• ➪Final number of moles in the gas n2 = m2/M = 1.5n1

• ➪Now eqn. of state of the gas in the initial state

• ➪P1V= n1 RT ---1.

➪Eqn. of state in the final state☟︎︎︎

• ➪P2V = n2 RT ---2.

• ➪By eqn. 1 and 2

• ➪P1/n1 = P2 /n2

• ➪= P2 = P1 (n2 /n1)

• ➪ P2 = 76 (1.5n1/n1)

• ➪P2 = 76x1.5

• ➪ P2 = 114 mm Hg

➪Requried Answer☟︎︎︎

• P2 = 114 mm Hg

Hope it helps you mate:)

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