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Nataly_w [17]
3 years ago
11

What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 23 ∘C , V1= 1.69×103 L ) to a v

olume of 3.38×103 L ? Assume no change in pressure or the amount of gas in the balloon.
Chemistry
1 answer:
andrezito [222]3 years ago
3 0

Answer:

319.15^{o}C[/tex]

Explanation:

When all other variables are constant, we are allowed to use the formula

\frac{T_{2} }{V_{2} } = \frac{T_{1} }{V_{1} } \\Which can be rewritten as T_{2} = \frac{T_{1} V_{2} }{V_{1} }if you make T2 the subject of the formula. This formula is true only if temperature is in Kelvin not degrees Celsius so T1 must be converted to KelvinNow to calculate T2[tex]T_{2}= \frac{296.15K*3.38.10^{3}L }{1.69.10^{3}L }= 592.3K = 319.15^{o}  C

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AleksAgata [21]
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<span>
<em>Explanation : </em>

Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where, 
P = Pressure of the gas (Pa)
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<span>
The given data for the gas </span></span>is,<span>
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V = </span>125 mL = 125 x 10⁻⁶ m³<span>
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R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?

By applying the formula,
103591 Pa x  </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
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</span>Moles (mol) = mass (g) / molar mass (g/mol)<span>

Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
   molar mass of the gas = mass / moles
                                          = 0.281 g / </span>3.90 x 10⁻³ mol
<span>                                          = 72.05 g/mol

</span>
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