At what temperature will a 1.0 g sample of neon gas exert a pressure of 500 torr in a 5.0 L container ?

or
At what temperature will a 0.0470 mol of a gas fill a balloon to 1.20 L under 0.988 atm pressure ?

PV=nRT

To find the temperature in both of these scenarios, we can use the ideal gas law, which states:

PV = nRT

Where:
P is the pressure
V is the volume
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin

We can rearrange the equation to solve for T:

T = PV / (nR)

Now, let's solve each scenario one by one:

1. Scenario: 1.0 g sample of neon gas exerting a pressure of 500 torr in a 5.0 L container.

Step 1: Convert the mass to moles.
To do this, we need to divide the given mass by the molar mass of neon, which is approximately 20.18 g/mol.

Moles = Mass / Molar mass
= 1.0 g / 20.18 g/mol
≈ 0.0495 mol

Step 2: Convert the pressure from torr to atm.
To do this, we need to divide the given pressure by 760 torr (since 1 atm = 760 torr).

Pressure (in atm) = 500 torr / 760 torr
≈ 0.6579 atm

Step 3: Plug the values into the equation.

T = (0.6579 atm) * (5.0 L) / (0.0495 mol * R)

Since R is the ideal gas constant, which has a value of 0.0821 L·atm/(mol·K), we can substitute it in.

T ≈ (0.6579 atm * 5.0 L) / (0.0495 mol * 0.0821 L·atm/(mol·K))

T ≈ 40.0 K

Therefore, at approximately 40.0 K, a 1.0 g sample of neon gas will exert a pressure of 500 torr in a 5.0 L container.

2. Scenario: 0.0470 mol of gas filling a balloon to 1.20 L under 0.988 atm pressure.

Step 1: Plug the values into the equation.

T = (0.988 atm * 1.20 L) / (0.0470 mol * R)

Using the same value for R as before (0.0821 L·atm/(mol·K)):

T ≈ (0.988 atm * 1.20 L) / (0.0470 mol * 0.0821 L·atm/(mol·K))

T ≈ 286.75 K

Therefore, at approximately 286.75 K, 0.0470 mol of gas will fill a balloon to 1.20 L under 0.988 atm pressure.

Please note that these calculations assume ideal behavior of gases and do not take into account any deviations or other factors that may affect the actual temperature.