AP Chemistry a week behind and I really need help :(
Please show work.
The ideal gas law is used to solve for an unknown variable for a given sample of gas. As in the ChemScope, pressure is measured in atmospheres (atm), volume is measured in liters (L), temperature is measured in Kelvins (K), and gas quantity is measured in moles (mol). If other units are given in a problem, they must first be converted. Use 0.0821 Latm/molK as the value for R, the ideal gas law constant.
A 2.0L sample of nitrogen gas, containing 0.15 mole N2, exerts what pressure at 25 °C?
How many moles of carbon dioxide (CO2) are contained within a 2500.mL vessel at 0.0 °C, given that the pressure is 1.15 atm?
Calculate the density of 2.0 moles ozone (O3) stored at room temperature (24 °C) under 1.5 atm of pressure. Think about how the formula for density could be obtained from the ideal gas law.
Please show work.
The ideal gas law is used to solve for an unknown variable for a given sample of gas. As in the ChemScope, pressure is measured in atmospheres (atm), volume is measured in liters (L), temperature is measured in Kelvins (K), and gas quantity is measured in moles (mol). If other units are given in a problem, they must first be converted. Use 0.0821 Latm/molK as the value for R, the ideal gas law constant.
A 2.0L sample of nitrogen gas, containing 0.15 mole N2, exerts what pressure at 25 °C?
How many moles of carbon dioxide (CO2) are contained within a 2500.mL vessel at 0.0 °C, given that the pressure is 1.15 atm?
Calculate the density of 2.0 moles ozone (O3) stored at room temperature (24 °C) under 1.5 atm of pressure. Think about how the formula for density could be obtained from the ideal gas law.

A 2.0L sample of nitrogen gas, containing 0.15 mole N2, exerts what pressure at 25 °C?
PV = nRT
P = ? atm
V = 2.0 L
n = 0.15 mol
R = 0.0821 L atm mol^1 K^1
T = 25 deg C = (25 + 273.15) K = 298.15 K
P = nRT / V
= 0.15 mol x 0.0821 x 298.15 K / 2.0 L
= 1.8 atm (2 sig figs)
How many moles of carbon dioxide (CO2) are contained within a 2500.mL vessel at 0.0 °C, given that the pressure is 1.15 atm?
P = 1.15 atm
V = 2500. ml = 2.500 L
n = ? mol
R = 0.0821 L atm mol^1 K^1
T = 0.0 deg C = 273.15 K
n = PV / RT
= 1.15 atm x 2.500 L / (0.0821 x 273.15)
= 0.128 mol (3 sig figs)
Calculate the density of 2.0 moles ozone (O3) stored at room temperature (24 °C) under 1.5 atm of pressure. Think about how the formula for density could be obtained from the ideal gas law
The ideal gas equation is PV = nRT
we know that moles (n) = mass (m) / molar mass (M)
this can be substituted into the ideal gas equation to give
PV = mRT / M
Now, density (d) = mass (m) / volume (V)
So if you rearrange the ideal gas equation now, to get m/V alone on one side then you will have the expression for density
PV = mRT / M
m/V = PM / RT
d = PM/RT
P = 1.5 atm
M = 48 g/mol
R = 0.0821 L atm mol^1 K^1
T = 297.15 K
d = 1.5 atm x 48 g/mol / (0.0821 x 297.15 K)
= 2.951 g/L
= 3.0 g/L (2 sig figs)
PV = nRT
P = ? atm
V = 2.0 L
n = 0.15 mol
R = 0.0821 L atm mol^1 K^1
T = 25 deg C = (25 + 273.15) K = 298.15 K
P = nRT / V
= 0.15 mol x 0.0821 x 298.15 K / 2.0 L
= 1.8 atm (2 sig figs)
How many moles of carbon dioxide (CO2) are contained within a 2500.mL vessel at 0.0 °C, given that the pressure is 1.15 atm?
P = 1.15 atm
V = 2500. ml = 2.500 L
n = ? mol
R = 0.0821 L atm mol^1 K^1
T = 0.0 deg C = 273.15 K
n = PV / RT
= 1.15 atm x 2.500 L / (0.0821 x 273.15)
= 0.128 mol (3 sig figs)
Calculate the density of 2.0 moles ozone (O3) stored at room temperature (24 °C) under 1.5 atm of pressure. Think about how the formula for density could be obtained from the ideal gas law
The ideal gas equation is PV = nRT
we know that moles (n) = mass (m) / molar mass (M)
this can be substituted into the ideal gas equation to give
PV = mRT / M
Now, density (d) = mass (m) / volume (V)
So if you rearrange the ideal gas equation now, to get m/V alone on one side then you will have the expression for density
PV = mRT / M
m/V = PM / RT
d = PM/RT
P = 1.5 atm
M = 48 g/mol
R = 0.0821 L atm mol^1 K^1
T = 297.15 K
d = 1.5 atm x 48 g/mol / (0.0821 x 297.15 K)
= 2.951 g/L
= 3.0 g/L (2 sig figs)