Answer:
[tex]W_{by}=9.00 \times 10^4 \ J\\\\W_{on}=-9.00 \times 10^4 \ J[/tex]
Option (c) is correct.
Explanation:
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Work done on/by a gas:}}\\W_{by}=P \Delta V \ or \ \int\limits^{V_f}_{V_0} {P} \, dV \\W_{on}=-P\Delta V \ or \ -\int\limits^{V_f}_{V_0} {P} \, dV \end{array}\right }[/tex]
Given:
[tex]P=1.8 \times 10^5 \ Pa\\\\V_0=1.2 \ m^3\\\\V_f=1.7 \ m^3[/tex]
Find:
[tex]W_{by}=?? \ J\\\\W_{on}=?? \ J[/tex]
(1) - Calculating the change in volume
[tex]\Delta V= V_f-V_0\\\\\Longrightarrow \Delta V=1.7-1.2\\\\\therefore \boxed{\Delta V=0.5 \ m^3}[/tex]
(2) - Calculating the work done by the gas
[tex]W_{by}=P \Delta V\\\\\Longrightarrow W_{by}=(1.8 \times 10^5)(0.5)\\\\\therefore \boxed{\boxed{W_{by}=9.00 \times 10^4 \ J}}[/tex]
(3) - Calculating the work done on the gas
[tex]W_{on}=-P \Delta V\\\\\Longrightarrow W_{on}=-(1.8 \times 10^5)(0.5)\\\\\therefore \boxed{\boxed{W_{on}=-9.00 \times 10^4 \ J}}[/tex]
Options (a) and (d) can be eliminated. Option (b) can be eliminated since there is no negative in front of the answer. This leaves the correct answer being option (c).