1. ans: 1/3; PV = nRT = constant, so P = const / V.

2. ans: 610 K; rearrange PV = nRT to T = PV/nR.

3. ans: 56; mass in grams = number of moles times molecular (or atomic) mass.

4. ans: 3.2 x10²⁴; = number of moles (=300/56) times N_A.

5. ans: 335 m/s; = sqrt[3RT /.071]. The .071 is the molar mass in kg (remember to have consistent SI units).

6. ans: 4.48 ATM; the pressure changes by the ratio of temperatures (in K) 523/283.

7. ans: 3.3 x10¹²; (10^-7/760 = no. ATMs pressure)*(1 liter) /RT = no. moles, * N_A.

8. ans: 310 K; = 37+273.

9. ans: 1200 moles; PV = nRT = 4 x10⁷ is a constant. Solve for n at both temperatures (in K): n=17623,16420. The difference is the answer (rounded to an appropriate number of significant figures).

10. ans: 200 øC; total mass of air inside balloon plus cargo mass (rho_hot*V + 200) must equal displaced mass rho_cold*V. Solve for rho_hot =0.75 kg/m³. Then, T_hot = T_cold *(rho_cold/rho_hot).

11. ans: 1.4 cm; the volume expansion factor is P_deep/P_surface = (rho_water*gh + 1 ATM)/(1 ATM) (all in SI units) =2.8. The diameter expansion factor is the cube root of the volume expansion.