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Nataliya [291]
3 years ago
12

A student increases the temperature of a 417cm³ balloon from 278k to 231k. Assuming constant pressure, what should the new volum

e of the balloon be?
A. 417cm³

B. 376cm³

C. 924cm³

D. 462cm³
Chemistry
1 answer:
viva [34]3 years ago
3 0
Charles law states that volume of gas is directly proportional to temperature at constant pressure 
V/T = k
where V - volume , T - temperature and k - constant
\frac{V1}{T1} =  \frac{V2}{T2}
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 

in the question it states that the temperature has been increased from 278 K to 231 K but it should actually be temperature is decreased from 278 K to 308 K
substituting the values in the equation 
\frac{417cm^{3} }{278K} =  \frac{V}{308 K}
V = 462 cm³
the answer should be D. 462 cm³ 
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Explanation:

1) <em>¿Cuál es la masa de agua a temperatura ambiente?</em>

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m_{o} = \left(1\,\frac{g}{mL} \right)\cdot (500\,mL)

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2) <em>¿Cuál es la masa de agua cuando se congela?</em>

Puesto que el proceso de congelación no implica transferencia de masa, la masa de agua se conserva al transformarse en hielo. Por tanto, la masa resultante es de 500 gramos.

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m_{f} =\rho_{w}\cdot V_{f}

Si sabemos que \rho_{w} = 1\,\frac{g}{mL} y V_{f} = 400\,mL, entonces:

m_{f} = \left(1\,\frac{g}{mL} \right)\cdot (400\,mL)

m_{f} = 400\,g

La masa de agua que queda después de la evaporación es de 400 gramos.

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m_{v} =m_{o}-m_{f}

Si m_{o} = 500\,g y m_{f} = 400\,g, entonces tenemos que:

m_{v} = 500\,g -400\,g

m_{v} = 100\,g

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