Question

An open vessel contains 200 mg of air at 17 degree C. What mass percent of air whole be expelled if the vessel is heated to 117 deg C ?

Answer 1

Answer : The mass percent of air whole be expelled, 25.5 %

Solution : Given,

Initial mass of air = 200 mg

Initial temperature = $17^oC=273+17=290K$

Final temperature = $117^oC=273+117=390K$

The ideal gas equation is,

$PV=nRT\\\\PV=\frac{w}{M}RT$

where,

P = pressure, V = volume, R = gas constant, T = temperature, w = mass and M = molar mass

The relation between the mass and temperature will be,

$\frac{T_1}{T_2}=\frac{w_2}{w_1}$

Now put all the given values in this relation, we get

$\frac{290K}{390K}=\frac{w_2}{200mg}$

$w_2=149mg$

Now we have to calculate the mass percent of air expelled.

$\text{Mass percent of air expelled}=\frac{200-149}{200}\times 100=25.5\%$

Therefore, the mass percent of air whole be expelled, 25.5 %

Answer 2

Explanation:

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