## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can calculate the work done on the gas: $W = -P\Delta V$ $W = -P ~(V_f-V_i)$ $W = -(4.00\times 10^5~Pa)(200\times 10^{-6}~m^3-600\times 10^{-6}~m^3)$ $W = 160~J$ The work done on the gas is 160 J. We can find the change in thermal energy of the gas as: $\Delta U = Q + W$ $\Delta U = (-100~J) + 160~J$ $\Delta U = 60~J$ The change in thermal energy of the gas is 60 J.