Question

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?

Answer 1

`pV = nRT`, or
`V = (nRT)/(p)`, with p - pressure, V - volume, n - amount of substance, T - temperature in K.

Then:
`(V_2)/(V_1) = (n_2RT_2p_1)/(p_2n_1RT_1)`

`n_2 = n_1`,
`p_2 = p_1`

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

Answer 2

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

Step-by-step 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