A balloon is inflated with 0.2494 g of helium to a pressure of 1.26 atm. If the desired volume of the balloon is 1.250 L, what must the temperature be in °C?

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A balloon is inflated with 0.2494 g of helium to a pressure of 1.26 atm. If the desired volume of the balloon is 1.250 L, what must the temperature be in °C?

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Answer 1 · 2016-06-04T04:56:38

Approx. $305 - 310 \cdot K$ $305-310*K$

Explanation:

We use the Ideal Gas equation and solve for $Temperature$ $"Temperature"$ :

$T = \frac{P V}{n R}$ $T=(PV)/(nR)$ $=$ $=$

$\frac{1.26 \cdot a t m \times 1.250 \cdot L}{\frac{0.2494 \cdot g}{4.003 \cdot g \cdot m o l^{- 1}} \times 0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1}}$ $(1.26*atmxx1.250*L)/((0.2494*g)/(4.003*g*mol^-1)xx0.0821*L*atm*K^-1*mol^-1)$ $=$ $=$ $? ? K$ $??K$

We get an answer in $Kelvin$ $"Kelvin"$ . We must convert this to $Centigrade$ $"Centigrade"$ .

$Centigrade Temperature$ $"Centigrade Temperature"$ $=$ $=$ $(Kelvin Temperature - 273.15)^{\circ} C$ $("Kelvin Temperature"-"273.15")""^@C$

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A balloon is inflated with 0.2494 g of helium to a pressure of 1.26 atm. If the desired volume of the balloon is 1.250 L, what must the temperature be in °C?

1 Answer

Explanation:

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