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A balloon filled with air has a volume of 4.24 liters at 23.00°
c. If the balloon is cooled at constant pressure to 5.00°C, what is its new volume?

Respuesta :

zjckbr

[tex] pV = nRT [/tex], or [tex] V = \frac{nRT}{p} [/tex], with p - pressure, V - volume, n - amount of substance, T - temperature in K.

Then: [tex] \frac{V_2}{V_1} = \frac{n_2RT_2p_1}{p_2n_1RT_1} [/tex]

[tex] n_2 = n_1 [/tex], [tex] p_2 = p_1 [/tex]

[tex] V_2 = V_1 \cdot \frac{T_2}{T_1} = 4.24 \frac {(273.00+5.00)}{(273.00+23.00)}=3.98 l [/tex]

kenmyna

The  new volume  if the balloon is cooled at constant pressure  is  3.98 L


 Explanation

The new  volume is calculated using the Charles law formula

that is V1/ T1 = V2/T2  where,

V1 = 4.24 l

T1= 23c  into kelvin = 23  +273 =296 K

T2 = 5.00 c into kelvin = 5.00 + 273 = 278 K

V2 = ?


by making  V2 subject the subject of the formula  by multiplying both side by T2

V2 = V1 xT2/ T1

V2 =  (4.24 L   x 278 K)  / 296 k  =3.98 L



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