Which one of these samples contains the smallest number of molecules?
a.
1.0 L of H2 at STP (0°C and 1atm)
d.
1.0 L of N2 at 0°C and 800 torr
b.
1.0 L of N2 at STP
e.
1.0 L of He at STP
c.
1.0 L of H2 at 20°C and 760 torr
I suppose the simplest thing to do is to calculate the number of mols of each although 1.0 L at STP will give the same number of mols no matter which gas is used.
PV = nRT
You can use reasoning if you wish. Note that n = PV/RT so when T is greater than standard T n will be greater and when P is larger than standard you know n will be larger. I figure the smallest one will be either c or d.
hjojp
To determine which sample contains the smallest number of molecules, 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.
At STP (standard temperature and pressure), the conditions are 0°C (273.15 K) and 1 atm.
a. 1.0 L of H2 at STP:
P = 1 atm, V = 1.0 L, T = 0°C = 273.15 K
Using the ideal gas law:
(1 atm)(1.0 L) = n(0.0821 L·atm/(mol·K))(273.15 K)
n = 0.0406 mol H2.
d. 1.0 L of N2 at 0°C and 800 torr:
P = 800 torr = 1.053 atm, V = 1.0 L, T = 0°C = 273.15 K
Using the ideal gas law:
(1.053 atm)(1.0 L) = n(0.0821 L·atm/(mol·K))(273.15 K)
n = 0.0432 mol N2.
b. 1.0 L of N2 at STP:
P = 1 atm, V = 1.0 L, T = 0°C = 273.15 K
Using the ideal gas law:
(1 atm)(1.0 L) = n(0.0821 L·atm/(mol·K))(273.15 K)
n = 0.0406 mol N2.
e. 1.0 L of He at STP:
P = 1 atm, V = 1.0 L, T = 0°C = 273.15 K
Using the ideal gas law:
(1 atm)(1.0 L) = n(0.0821 L·atm/(mol·K))(273.15 K)
n = 0.0406 mol He.
c. 1.0 L of H2 at 20°C and 760 torr:
P = 760 torr = 1 atm, V = 1.0 L, T = 20°C = 293.15 K
Using the ideal gas law:
(1 atm)(1.0 L) = n(0.0821 L·atm/(mol·K))(293.15 K)
n = 0.0458 mol H2.
Comparing the number of moles in each sample:
a. H2 at STP: 0.0406 mol
d. N2 at 0°C and 800 torr: 0.0432 mol
b. N2 at STP: 0.0406 mol
e. He at STP: 0.0406 mol
c. H2 at 20°C and 760 torr: 0.0458 mol
Therefore, sample a (1.0 L of H2 at STP) contains the smallest number of molecules with 0.0406 mol.
To determine which sample contains the smallest number of molecules, we need to compare the conditions of each sample and use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of molecules (in moles), R is the ideal gas constant, and T is the temperature in Kelvin.
Let's examine each sample individually:
a. 1.0 L of H2 at STP (0°C and 1 atm)
At STP, the temperature is 0°C, which is equivalent to 273 K. The pressure is 1 atm, and the volume is 1.0 L. Using the ideal gas law equation, we have:
(1 atm)(1.0 L) = n(0.0821 L·atm/mol·K)(273 K)
n = 1.0 mol
d. 1.0 L of N2 at 0°C and 800 torr
The temperature is 0°C, which is 273 K. The pressure is given in torr and needs to be converted to atm: 800 torr ÷ 760 torr/atm = 1.053 atm. The volume is 1.0 L. Using the ideal gas law equation, we have:
(1.053 atm)(1.0 L) = n(0.0821 L·atm/mol·K)(273 K)
n = 1.049 mol
b. 1.0 L of N2 at STP
At STP, the temperature is 0°C, which is 273 K. The pressure is 1 atm, and the volume is 1.0 L. Using the ideal gas law equation, we have:
(1 atm)(1.0 L) = n(0.0821 L·atm/mol·K)(273 K)
n = 1.0 mol
e. 1.0 L of He at STP
At STP, the temperature is 0°C, which is 273 K. The pressure is 1 atm, and the volume is 1.0 L. Using the ideal gas law equation, we have:
(1 atm)(1.0 L) = n(0.0821 L·atm/mol·K)(273 K)
n = 1.0 mol
c. 1.0 L of H2 at 20°C and 760 torr
The temperature is 20°C, which is 293 K. The pressure is 760 torr ÷ 760 torr/atm = 1 atm. The volume is 1.0 L. Using the ideal gas law equation, we have:
(1 atm)(1.0 L) = n(0.0821 L·atm/mol·K)(293 K)
n = 1.0 mol
By comparing the number of moles for each sample, we can see that samples a, b, e, and c all have the same number of moles, which is 1.0 mol. However, sample d has a slightly higher number of moles, 1.049 mol. Therefore, sample d contains the smallest number of molecules.