Respuesta :
Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 2 L, [tex]V_{2}[/tex] = 3 L,
[tex]P_{1}[/tex] = X, [tex]P_{2}[/tex] = 0.1 X,
[tex]m_{1}[/tex] = 4.8 g, [tex]m_{2}[/tex] = 0.36 g,
[tex]M_{1}[/tex] = ?, [tex]M_{2}[/tex] = ?
According to the ideal gas equation,
[tex]\frac{P_{1}V_{1}M_{1}}{m_{1}} = \frac{P_{2}V_{2}M_{2}}{m_{2}}[/tex]
[tex]\frac{X \times 2 \times M_{1}}{4.8 g} = \frac{0.1X \times 3 \times M_{2}}{0.36 g}[/tex]
[tex]\frac{M_{1}}{M_{2}} = \frac{4.8 \times 0.1 \times 3}{2 \times 0.36}[/tex]
= 2
or, [tex]M_{1} = 2 \times M_{2}[/tex]
Thus, we can conclude that two gases does not have the same molar mass and flask 1 contains a gas of higher molar mass.
The two gases do not have a similar molar mass and flask 1 comprise a gas of higher molar mass.
Calculation of the molar mass:
Since volume 2 L and the other of volume 3 L. The 2-L flask contains 4.8 g of gas, and the gas pressure is X atm. The 3-L flask contains 0.36 g of gas, and the gas pressure is 0.1X.
Now
x*2 *m1 /4.8g = 0.1x * 3 * m2/0.36g
m1/m2 = 4.8*0.1*3 \2*0.36
= 2
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