## Essential University Physics: Volume 1 (3rd Edition)

a) $V_f = 3351 \ m^3$ b) $T=7.815^{\circ}C$
a) This is an adiabatic process. We know that for adiabatic processes, $PV^{\gamma}$ is constant. Also, for Helium, we know that the value of $\gamma$ is 1.66. Thus, we find: $(1)((1.75\times10^3)^{1.66})=(.34)(V^{1.66}) \\ V_f = 3351 \ m^3$ b) We use the equation for temperature for an adiabatic process, which is equation 18.11b in the textbook. Doing this, we find: $(12)(1.75\times10^3)^{.66}=T(3351)^{.66}$ $T=7.815^{\circ}C$