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Question: In a process, the volume of gas increases from \[1000\,{\text{cc}}\] to \[2000\,{\text{cc}}\] at con...

In a process, the volume of gas increases from 1000 cc1000\,{\text{cc}} to 2000 cc2000\,{\text{cc}} at constant pressure 100 kPa100\,{\text{kPa}}. Find work done by the gas.

Explanation

Solution

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 WW by the gas is given by
W=PΔVW = P\Delta V …… (1)
Here, PP is pressure of the gas and ΔV\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 cc1000\,{\text{cc}} to 2000 cc2000\,{\text{cc}}. Hence, the initial volume of the gas is 1000 cc1000\,{\text{cc}} and the final volume of the gas is 2000 cc2000\,{\text{cc}}.
vi=1000 cc{v_i} = 1000\,{\text{cc}}
⇒vf=2000 cc\Rightarrow {v_f} = 2000\,{\text{cc}}
The constant pressure of the gas is 100 kPa100\,{\text{kPa}}.
P=100 kPaP = 100\,{\text{kPa}}
Convert the unit of initial volume of the gas in the SI system of units.
vi=(1000 cc)(10−6 m31 cc){v_i} = \left( {1000\,{\text{cc}}} \right)\left( {\dfrac{{{{10}^{ - 6}}\,{{\text{m}}^3}}}{{1\,{\text{cc}}}}} \right)
⇒vi=10−3 m3\Rightarrow {v_i} = {10^{ - 3}}\,{{\text{m}}^3}
Hence, the initial volume of the gas is 10−3 m3{10^{ - 3}}\,{{\text{m}}^3}.

Convert the unit of final volume of the gas in the SI system of units.
vf=(2000 cc)(10−6 m31 cc){v_f} = \left( {2000\,{\text{cc}}} \right)\left( {\dfrac{{{{10}^{ - 6}}\,{{\text{m}}^3}}}{{1\,{\text{cc}}}}} \right)
⇒vf=2×10−3 m3\Rightarrow {v_f} = 2 \times {10^{ - 3}}\,{{\text{m}}^3}
Hence, the final volume of the gas is 2×10−3 m32 \times {10^{ - 3}}\,{{\text{m}}^3}.
Convert the unit of pressure of the gas in the SI system of units.
P=(100 kPa)(103 Pa1 kPa)P = \left( {100\,{\text{kPa}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{Pa}}}}{{1\,{\text{kPa}}}}} \right)
⇒P=105 Pa\Rightarrow P = {10^5}\,{\text{Pa}}
Hence, the pressure of the gas is 105 Pa{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(Vf−Vi)W = P\left( {{V_f} - {V_i}} \right)
Substitute 105 Pa{10^5}\,{\text{Pa}} for PP, 2×10−3 m32 \times {10^{ - 3}}\,{{\text{m}}^3} for Vf{V_f} and 10−3 m3{10^{ - 3}}\,{{\text{m}}^3} for Vi{V_i} in the above equation.
W=(105 Pa)(2×10−3 m3−10−3 m3)W = \left( {{{10}^5}\,{\text{Pa}}} \right)\left( {2 \times {{10}^{ - 3}}\,{{\text{m}}^3} - {{10}^{ - 3}}\,{{\text{m}}^3}} \right)
⇒W=(105 Pa)(1×10−3 m3)\Rightarrow W = \left( {{{10}^5}\,{\text{Pa}}} \right)\left( {1 \times {{10}^{ - 3}}\,{{\text{m}}^3}} \right)
⇒W=102 J\Rightarrow W = {10^2}\,{\text{J}}
∴W=100 J\therefore W = 100\,{\text{J}}

Hence, the work done by the gas is 100 J100\,{\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.