Answer to Question #109454 in General Chemistry for Nate

Question

1. At an ocean depth of 10.0 m, a diver’s lung capacity is 2.40 L. The air temperature is 32.0°C and the pressure is 101.30 kPa. What is the volume of the diver’s lungs at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa? 2. In a mixture of gases the partial pressure of nitrogen is 0.79 atm, the partial pressure of oxygen is 0.20 atm, and the partial pressure of hydrogen is 0.00444 atm. What is the total pressure of the mixture of gases? 3. Which law describes the relationship between the pressure, volume, and temperature of a gas? 4. What volume does 0.0685 mol of a gas occupy at STP? 5. If I have an unknown quantity of gas at a pressure of 1.2 atm, a volume of 31 liters, and a temperature of 87°C, how many moles of the gas do I have?

Answer 1

V = 1.66 L

Explanation:

As we know that at the depth of 10 m the volume of the lungs of diver is given as

V = 2.40 L

Temperature = 32 degree C

Pressure = 101.30 kPa

Now we have change the values of pressure and temperature

Pressure = 141.20 kPa

Temperature = 21 degree C

Now we can use idea gas equation to solve this

so we have