Answer
a. $4.2\times10^5J$
b. $6.4\times10^5J $
Work Step by Step
a. Calculate the work done by a gas at constant pressure from equation 15–3.
$$W=P\Delta V=(1atm)(\frac{1.01\times10^5Pa}{1atm})(16.2m^3-12.0m^3)=4.2\times10^5J$$
b. Calculate the change in internal energy from $\Delta U=Q-W$
$$\delta U = (254kcal)\frac{4186J}{1kcal}-4.242\times10^5J=6.4\times10^5J $$
