Question

A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained
the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.)
O 0.7g
O 1.4g
O 0.675g
O 0.7 mol​

Answer 1

Given :

• Initial mass of Helium = 2g
• Initial volume of helium = 2L
• Final volume of helium = 2.7L
• Pressure = constant
• Temperature = Constant

To find :

Amount of helium added

Solution :

We know that, PV= nRT

Here we know that R is constant and P,T are also constant, this gives that , V ∝ n

$\sf \implies \dfrac{V_1}{V_2} = \dfrac{n_1}{n_2}$

We know that , n = number of mole = mass/molar mass

$\sf \implies \dfrac{V_1}{V_2} = \dfrac{ \dfrac{initial \: mass}{molar \: mass} }{ \dfrac{final \: mass}{molar \: mass} }$

$\sf \implies \dfrac{V_1}{V_2} = \dfrac{\: initial \: mass}{ final \: mass}$

$\sf \implies \dfrac{2}{2.7} = \dfrac{2}{final \: mass}$

$\sf \implies \: {final \: mass}\: = \dfrac{2.7}{2} \times 2$

$\sf \implies \: {final \: mass}\: = \: 2.7g$

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Also,

➝ final mass = initial mass + amount of helium added

➩ 2.7g = 2g + Amount of helium added

➩ Amount of helium added = 2.7g - 2g

➩ Amount of helium added = 0.7g

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ANSWER :

Option 1) 0.7 g