202203071449 Exercises 13.1

A given mass of gas has a volume of 144\,\mathrm{cm^3} at 15\mathrm{^\circ C}. Calculate its volume at i. 33\mathrm{^\circ C}, ii. 0\mathrm{^\circ C}, iii. -67\mathrm{^\circ C}, the pressure being constant.

Extracted from M. Nelkon. (1971). Graded Exercises and Worked Examples in Physics.


Solution.

Converting cubic centimetres to cubic metres:

144\,\mathrm{cm^3}=144\,\mathrm{(0.01\,m)^3}=0.000144\,\mathrm{m^3}.

Converting degree Celsius to Kelvin:

15\mathrm{^\circ C}=(15+273.15)\,\mathrm{K}=288.15\,\mathrm{K}.


Background. (ideal gas equation)

By ideal gas law PV=nRT, where P, V, and T are the pressure, volume, and temperature; n is the amount of substance; and R is the ideal gas constant.

Wikipedia on Ideal gas law


Arranging pV=nRT for change of subject,

\displaystyle{V=\bigg( \frac{nR}{p}\bigg) T}.

Then plugging in V=0.000144 and T=288.15, one obtains

\begin{aligned} 0.000144 & = \bigg( \frac{nR}{p} \bigg) (288.15) \\ \frac{nR}{p} & = \frac{0.000144}{288.15} \textrm{ ( = constant)} \\ \end{aligned}

Thus we have volume V as a function V(T) of temperature T:

V(T)=(4.9974\times 10^{-7})\times T.

We are asked about its volume as the temperature varies:

\begin{aligned} 33\mathrm{^\circ C} & \Leftrightarrow (33+273.15)\,\mathrm{K} = 306.15\,\mathrm{K} \\ 0\mathrm{^\circ C} & \Leftrightarrow (0+273.15)\,\mathrm{K} = 273.15\,\mathrm{K} \\ -67\mathrm{^\circ C} & \Leftrightarrow (-67+273.15)\,\mathrm{K} = 206.15\,\mathrm{K} \\ \end{aligned}

(to be continued)