Problem involving combined gas law
A scuba diver is preparing for a deep dive in the ocean. The diver's tank has a volume of 10 liters and contains a mixture of oxygen and nitrogen gases. The temperature of the tank is initially 20 degrees Celsius and the pressure is 2 atm.
During the dive, the temperature of the tank increases to 30 degrees Celsius. What is the new pressure in the tank?
To solve this problem, we can use the combined gas law, which states:
\(P_1V_1/T_1 = P_2V_2/T_2\)
Where:
\(P_1 = initial pressure = 2 atm\)
\(V_1 = initial volume = 10 liters\)
\(T_1 = initial temperature = 20°C + 273 = 293 K\)
\(P_2 = final pressure = ?\)
\(V_2 = same volume = 10 liters\)
\(T_2 = final temperature = 30°C + 273 = 303 K\)
Now we can plug in the values and solve for the final pressure:
\(2 * 10 / 293 = P_2 * 10 / 303\)
\(20 / 293 = P_2 / 30.3\)
\(P_2 = (20 / 293) * 30.3\)
\(P_2 ≈ 2.07 atm\)
Therefore, the new pressure in the tank after the temperature increase is approximately 2.07 atm.