What is the temperature of 2.48 moles of gas stored in a 30.0 L container at 1559 mmHg
PV = nRT
P = 1559 mm/760 = ? atm
V= 30.0 L
n = 2.48 moles
R = 0.08205 L*atm/mol*K
T = ? kelvin. If you want T in celsius then Kelvin = 273.15 + Celsius.
Post your work if you get stuck.
To find the temperature of a gas, we can use the ideal gas law, which states that PV = nRT, where:
- P represents the pressure of the gas,
- V represents the volume of the gas,
- n represents the number of moles of gas,
- R is the ideal gas constant, and
- T represents the temperature in Kelvin.
First, we need to convert the pressure from mmHg to atm, as the ideal gas constant has units in atm:
1 atm = 760 mmHg
1559 mmHg ÷ 760 mmHg/atm = 2.05 atm
Now we can rearrange the ideal gas law equation to solve for T:
T = PV / nR
Plugging in the given values:
P = 2.05 atm
V = 30.0 L
n = 2.48 moles
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = (2.05 atm * 30.0 L) / (2.48 moles * 0.0821 L·atm/(mol·K))
T ≈ 96.81 K
Therefore, the temperature of 2.48 moles of gas stored in a 30.0 L container at 1559 mmHg is approximately 96.81 K.
To find the temperature of the gas, 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.
In this case, we are given the pressure (1559 mmHg), the volume (30.0 L), and the number of moles (2.48 moles). We can use these values to solve for the temperature.
First, we need to convert the pressure from mmHg to atm. Since 1 atm = 760 mmHg, we divide the given pressure by 760:
1559 mmHg / 760 mmHg/atm = 2.05 atm (rounded to two decimal places)
Next, we need to rearrange the ideal gas law equation to solve for temperature, which gives us:
T = PV / nR
Now we can substitute the known values:
T = (2.05 atm) x (30.0 L) / (2.48 moles) x (0.0821 L·atm/mol·K) (Note: R is the ideal gas constant)
We can cancel the units to simplify the equation:
T = (2.05 x 30.0) / (2.48 x 0.0821) K
T ≈ 75.13 K
Therefore, the temperature of 2.48 moles of gas stored in a 30.0 L container at 1559 mmHg is approximately 75.13 K.