Question

Calculate the fall in temperature of helium initially at 15°C when it is suddenly expanded to 8 times its original volume (𝛾 = 5/3). 

Answer 1

Given:

T_i = 15°C = 15 + 273 = 288 K

𝛾 = `5/3` , V_f = 8 V_i 

To find: Fall in temperature (ΔT)

Formulae: `"T"_"f""V"_"f"^{Υ - 1} = "T"_"i""V"_"i"^"Υ - 1"`

Calculation:

From formula, 

`"T"_"f" = "T"_"i"("V"_"i"/"V"_"f")^"Υ - 1"`

= `288(1/8)^{5/3 - 1}`

= `288(1/8)^(2/3)`

= `288 xx 1/4`

= 72 K

= 72 - 273

= -201°C

∴ ΔT = T_f - T_i 

= -201 - 15

= -216°C

Fall in the temperature (ΔT) is -216°C.