## College Physics (4th Edition)

$1.34\times 10^{26}$ molecules were released.
We can find the original number of molecules in the tank: $P~V = N_1~k~T$ $N_1 = \frac{P~V}{k~T}$ $N_1 = \frac{(20.0\times 1.01\times 10^5~Pa)(1.0~m^3)}{(1.38\times 10^{-23}~J/K)(273~K)}$ $N_1 = 5.36\times 10^{26}$ We can find the new number of molecules in the tank after the valve is opened: $P~V = N_2~k~T$ $N_2 = \frac{P~V}{k~T}$ $N_2 = \frac{(15.0\times 1.01\times 10^5~Pa)(1.0~m^3)}{(1.38\times 10^{-23}~J/K)(273~K)}$ $N_2 = 4.02\times 10^{26}$ We can find the number of molecules that were released: $N_1-N_2 = (5.36\times 10^{26}) - (4.02\times 10^{26}) = 1.34\times 10^{26}$ $1.34\times 10^{26}$ molecules were released.