1. Calculate the number of molecules in a deep breath of air whose volume is 2.15 L at body temperature, 36 degrees Celsius, and a pressure of 740 torr.

2.A fixed quantity of gas at 24 degrees Celsius exhibits a pressure of 740 torr and occupies a volume of 4.42 L.
a)Use Boyle's law to calculate the volume the gas will occupy if the pressure is increased to 1.85 atm while the temp is held constant.
b)Use Charles's law to calculate the volume the gas will occupy if the temperature is increased to 160 degrees Celsius while the pressure is held constant.

3.A scuba diver's tank contains 0.29 kg of O2 compressed into a volume of 2.3 L.
a)Calculate the gas pressure inside the tank at 9 degrees Celsius.
b)What volume would this oxygen occupy at 25 degrees Celsius and 0.94 atm?

and
4. Calculate the rms speed of NF3 molecules at 21 degrees Celsius.
Somebody please help me!

1) 4.98*10^23

234

1. To calculate the number of molecules in a deep breath of air, we can use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, convert the given temperature from Celsius to Kelvin:
T(K) = T(C) + 273.15
T(K) = 36 + 273.15 = 309.15 K

Next, convert the given volume from liters to moles:
V(mol) = V(L) / 22.4
V(mol) = 2.15 / 22.4 = 0.096 mol

Now, rearrange the Ideal Gas Law equation to solve for n:
n = (PV) / (RT)
n = (740 torr * 2.15 L) / (0.0821 L·atm/mol·K * 309.15 K)
n ≈ 0.212 mol

Finally, calculate the number of molecules using Avogadro's number:
Number of molecules = n * Avogadro's number
Number of molecules = 0.212 mol * 6.022 x 10^23 molecules/mol
Number of molecules ≈ 1.276 x 10^23 molecules

Therefore, the number of molecules in a deep breath of air with a volume of 2.15 L at body temperature and a pressure of 740 torr is approximately 1.276 x 10^23 molecules.

2a) Boyle's law states that at constant temperature, the pressure and volume of a gas are inversely proportional. The equation for Boyle's law is P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Given:
P1 = 740 torr
V1 = 4.42 L
P2 = 1.85 atm (convert to torr: 1 atm = 760 torr)
T = 24 degrees Celsius (convert to Kelvin: T(K) = T(C) + 273.15)

Rearrange the equation to solve for V2:
V2 = (P1 * V1) / P2
V2 = (740 torr * 4.42 L) / (1.85 atm * 760 torr/atm)
V2 ≈ 10.57 L

Therefore, the gas will occupy a volume of approximately 10.57 L if the pressure is increased to 1.85 atm while the temperature is held constant.

2b) Charles's law states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature. The equation for Charles's law is V1 / T1 = V2 / T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.

Given:
V1 = 4.42 L
T1 = 24 degrees Celsius (convert to Kelvin: T(K) = T(C) + 273.15)
T2 = 160 degrees Celsius (convert to Kelvin: T(K) = T(C) + 273.15)
P = 740 torr

Rearrange the equation to solve for V2:
V2 = (V1 * T2) / T1
V2 = (4.42 L * (160 + 273.15) K) / (24 + 273.15) K
V2 ≈ 31.68 L

Therefore, the gas will occupy a volume of approximately 31.68 L if the temperature is increased to 160 degrees Celsius while the pressure is held constant.

3a) To calculate the gas pressure inside the scuba diver's tank, we can use the formula PV = nRT.

Given:
m = 0.29 kg (convert to grams: m(g) = m(kg) * 1000 g/kg)
V = 2.3 L
T = 9 degrees Celsius (convert to Kelvin: T(K) = T(C) + 273.15)

Convert the mass to moles using the molar mass of O2:
n = (m(g) / Molar mass of O2)
n = (0.29 kg * 1000 g/kg) / 32g/mol
n ≈ 9.0625 mol

Rearrange the equation to solve for P:
P = (nRT) / V
P = (9.0625 mol * 0.0821 L·atm/mol·K * (9 + 273.15) K) / 2.3 L
P ≈ 1046.28 atm (convert to torr: 1 atm = 760 torr)

Therefore, the gas pressure inside the scuba diver's tank at 9 degrees Celsius is approximately 1046.28 torr.

3b) To calculate the volume the oxygen would occupy at a different temperature and pressure, we can use the formula PV = nRT.

Given:
T1 = 9 degrees Celsius (convert to Kelvin: T1(K) = T1(C) + 273.15)
T2 = 25 degrees Celsius (convert to Kelvin: T2(K) = T2(C) + 273.15)
P = 0.94 atm
V1 = 2.3 L

Rearrange the equation to solve for V2:
V2 = (nR * T2) / P
V2 = (9.0625 mol * 0.0821 L·atm/mol·K * (25 + 273.15) K) / 0.94 atm
V2 ≈ 688.63 L

Therefore, the oxygen would occupy a volume of approximately 688.63 L at 25 degrees Celsius and 0.94 atm.

4. To calculate the rms speed of NF3 molecules at 21 degrees Celsius, we can use the formula vrms = √(3RT / M), where R is the ideal gas constant, T is the temperature in Kelvin, and M is the molar mass of NF3.

Given:
T = 21 degrees Celsius (convert to Kelvin: T(K) = T(C) + 273.15)
Molar mass of NF3 ≈ 71 g/mol

Convert the molar mass from grams to kilograms:
M = Molar mass of NF3 / 1000
M = 71 g/mol / 1000 = 0.071 kg/mol

Rearrange the equation to solve for vrms:
vrms = √(3RT / M)
vrms = √(3 * 0.0821 L·atm/mol·K * (21 + 273.15) K) / 0.071 kg/mol
vrms ≈ 533.78 m/s

Therefore, the rms speed of NF3 molecules at 21 degrees Celsius is approximately 533.78 m/s.