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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?

Answer 1

Caleb Adams

Approx. #305-310*K#

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

#T=(PV)/(nR)# #=#

#(1.26*atmxx1.250*L)/((0.2494*g)/(4.003*g*mol^-1)xx0.0821*L*atm*K^-1*mol^-1)# #=# #??K#

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

#"Centigrade Temperature"# #=# #("Kelvin Temperature"-"273.15")""^@C#

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Answer 2

Olivia Anton

Use the Ideal Gas Law: (PV = nRT). Convert grams of helium to moles: (n = \frac{0.2494 , \text{g}}{4.0026 , \text{g/mol}}). Rearrange the equation to solve for temperature: (T = \frac{PV}{nR}). Substitute values and solve.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.