Question

In a process, the volume of gas increases from \[1000\,{\text{cc}}\] to \[2000\,{\text{cc}}\] at constant pressure \[100\,{\text{kPa}}\]. Find work done by the gas.

Answer 1

Hint:We can the formula for pressure volume work done by the gas. This formula gives the relation between the pressure of the gas and change in volume of the gas. We can convert the units of initial and final volumes of the gas and pressure of the gas in the SI system of units and then substitute all these values in the formula to calculate work done by the gas.

Formula used:
The formula for the pressure volume work done \[W\] by the gas is given by
\[W = P\Delta V\] …… (1)
Here, \[P\] is pressure of the gas and \[\Delta V\] is change in volume of the gas.

Complete step by step answer:
We have given that the volume of gas in the process increases from \[1000\,{\text{cc}}\] to \[2000\,{\text{cc}}\]. Hence, the initial volume of the gas is \[1000\,{\text{cc}}\] and the final volume of the gas is \[2000\,{\text{cc}}\].
\[{v_i} = 1000\,{\text{cc}}\]
\[\Rightarrow {v_f} = 2000\,{\text{cc}}\]
The constant pressure of the gas is \[100\,{\text{kPa}}\].
\[P = 100\,{\text{kPa}}\]
Convert the unit of initial volume of the gas in the SI system of units.
\[{v_i} = \left( {1000\,{\text{cc}}} \right)\left( {\dfrac{{{{10}^{ - 6}}\,{{\text{m}}^3}}}{{1\,{\text{cc}}}}} \right)\]
\[ \Rightarrow {v_i} = {10^{ - 3}}\,{{\text{m}}^3}\]
Hence, the initial volume of the gas is \[{10^{ - 3}}\,{{\text{m}}^3}\].

Convert the unit of final volume of the gas in the SI system of units.
\[{v_f} = \left( {2000\,{\text{cc}}} \right)\left( {\dfrac{{{{10}^{ - 6}}\,{{\text{m}}^3}}}{{1\,{\text{cc}}}}} \right)\]
\[ \Rightarrow {v_f} = 2 \times {10^{ - 3}}\,{{\text{m}}^3}\]
Hence, the final volume of the gas is \[2 \times {10^{ - 3}}\,{{\text{m}}^3}\].
Convert the unit of pressure of the gas in the SI system of units.
\[P = \left( {100\,{\text{kPa}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{Pa}}}}{{1\,{\text{kPa}}}}} \right)\]
\[ \Rightarrow P = {10^5}\,{\text{Pa}}\]
Hence, the pressure of the gas is \[{10^5}\,{\text{Pa}}\].

Let us now calculate the work done by the gas.Rewrite equation (1) for the work done by the gas.
\[W = P\left( {{V_f} - {V_i}} \right)\]
Substitute \[{10^5}\,{\text{Pa}}\] for \[P\], \[2 \times {10^{ - 3}}\,{{\text{m}}^3}\] for \[{V_f}\] and \[{10^{ - 3}}\,{{\text{m}}^3}\] for \[{V_i}\] in the above equation.
\[W = \left( {{{10}^5}\,{\text{Pa}}} \right)\left( {2 \times {{10}^{ - 3}}\,{{\text{m}}^3} - {{10}^{ - 3}}\,{{\text{m}}^3}} \right)\]
\[ \Rightarrow W = \left( {{{10}^5}\,{\text{Pa}}} \right)\left( {1 \times {{10}^{ - 3}}\,{{\text{m}}^3}} \right)\]
\[ \Rightarrow W = {10^2}\,{\text{J}}\]
\[ \therefore W = 100\,{\text{J}}\]

Hence, the work done by the gas is \[100\,{\text{J}}\].

Note:The students should keep in mind that the initial and final volumes of the gas in the given process are in the CGS system of units. The students should not forget to convert units of these volumes in the SI system of units. Also the constant pressure of the gas is not in the SI system of units. The unit of pressure should also be converted from kilopascal to Pascal.