Ideal Gas Law
The ideal gas law is a fundamental principle in the field of thermodynamics that describes the behavior of gases at different temperatures, pressures, and volumes. It is represented mathematically by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature.
Here are some examples of how the ideal gas law can be applied:
A balloon is filled with helium gas at a temperature of 25°C and a pressure of 1 atm. If the balloon has a volume of 5 liters, how many moles of helium gas are inside?
PV = nRT
n = (PV) / (RT)
n = (1 atm x 5 L) / (0.0821 L atm/mol K x 298 K)
n = 0.204 moles
A scuba diver's tank contains 10 liters of air at a pressure of 200 atm and a temperature of 20°C. What is the volume of the air when it is released into the water at a depth of 40 meters, where the pressure is 300 atm and the temperature is 10°C?
PV = nRT
V1 = (nRT1) / P1
V1 = (n x 293 K x 200 atm) / (1 atm)
V1 = 58.65 L
V2 = (nRT2) / P2
V2 = (n x 283 K x 300 atm) / (1 atm)
V2 = 63.22 L
The volume of the air at the new conditions can be calculated using the combined gas law: P1V1/T1 = P2V2/T2
V2 = (P1V1T2) / (T1P2)
V2 = (200 atm x 58.65 L x 283 K) / (293 K x 300 atm)
V2 = 38.84 L
Therefore, the volume of the air at the new conditions is 38.84 liters.
A gas sample is compressed from a volume of 5 L to a volume of 2 L at a constant temperature of 25°C. If the pressure of the gas is initially 1 atm, what is the pressure after the compression?
PV = nRT
P1V1 = nRT
P2V2 = nRT
Since the temperature is constant, we can set the two equations equal to each other and solve for P2:
P1V1 = P2V2
P2 = (P1V1) / V2
P2 = (1 atm x 5 L) / 2 L
P2 = 2.5 atm
Therefore, the pressure of the gas after the compression is 2.5 atm.
A sample of oxygen gas has a volume of 2 L at a pressure of 3 atm and a temperature of 25°C. If the temperature is increased to 50°C, what is the new pressure of the gas?
PV = nRT
P1V1/T1 = P2V2/T2
We can solve for P2:
P2 = (P1V1T2) / (V2T1)
P2 = (3 atm x 2 L x 323 K) / (2 L x 298 K)
P2 = 3.46 atm
Therefore, the new pressure of the oxygen gas is 3.46 atm.
A sample of carbon dioxide gas has a volume of 1 L at a pressure of 2 atm and a temperature of 20°C. If the pressure is increased to 3 atm, what is the new volume of the gas?
PV = nRT
P1V1/T1 = P2V2/T2
We can solve for V2:
V2 = (P1V1T2) / (P2T1)
V2 = (2 atm x 1 L x 293 K) / (3 atm x 293 K)
V2 = 0.67 L
Therefore, the new volume of the carbon dioxide gas is 0.67 L.
A sample of nitrogen gas has a volume of 4 L at a pressure of 1 atm and a temperature of 25°C. If the volume is decreased to 2 L at a constant temperature, what is the new pressure of the gas?
PV = nRT
P1V1/T1 = P2V2/T2
We can solve for P2:
P2 = (P1V1T1) / (V2T2)
P2 = (1 atm x 4 L x 298 K) / (2 L x 298 K)
P2 = 2 atm
Therefore, the new pressure of the nitrogen gas is 2 atm.
These examples illustrate how the ideal gas law can be used to calculate different properties of gases when their temperature, pressure, and/or volume change.
A gas is contained in a piston with a movable wall. The initial volume is 1 L, and the pressure is 2 atm. If the pressure is increased to 4 atm while the temperature remains constant, what is the new volume of the gas?
PV = nRT
P1V1 = nRT
P2V2 = nRT
Since the temperature is constant, we can set the two equations equal to each other and solve for V2:
P1V1 = P2V2
V2 = (P1V1) / P2
V2 = (2 atm x 1 L) / 4 atm
V2 = 0.5 L
Therefore, the new volume of the gas is 0.5 L.
A sample of hydrogen gas has a pressure of 1 atm and a temperature of 0°C. If the gas is heated to 100°C while the volume remains constant, what is the new pressure of the gas?
PV = nRT
P1V1/T1 = P2V2/T2
Since the volume is constant, we can simplify the equation to:
P1/T1 = P2/T2
We can solve for P2:
P2 = (P1 x T2) / T1
P2 = (1 atm x 373 K) / 273 K
P2 = 1.37 atm
Therefore, the new pressure of the hydrogen gas is 1.37 atm.
A gas occupies a volume of 10 L at a pressure of 2 atm and a temperature of 25°C. If the volume is reduced to 5 L while the pressure remains constant, what is the new temperature of the gas?
PV = nRT
P1V1/T1 = P2V2/T2
Since the pressure is constant, we can simplify the equation to:
V1/T1 = V2/T2
We can solve for T2:
T2 = (V2 x T1) / V1
T2 = (5 L x 298 K) / 10 L
T2 = 149.4 K
Therefore, the new temperature of the gas is 149.4 K.
A sample of air has a volume of 5 L at a pressure of 1 atm and a temperature of 25°C. If the gas is compressed to a volume of 2 L at a constant temperature, what is the new pressure of the gas?
PV = nRT
P1V1/T1 = P2V2/T2
Since the temperature is constant, we can simplify the equation to:
P1V1 = P2V2
We can solve for P2:
P2 = (P1 x V1) / V2
P2 = (1 atm x 5 L) / 2 L
P2 = 2.5 atm
Therefore, the new pressure of the air is 2.5 atm.
This example shows how the ideal gas law can be used to calculate the new pressure of a gas when it is compressed to a smaller volume while the temperature remains constant.
These examples demonstrate how the ideal gas law can be used to calculate different properties of gases when their temperature, pressure, and/or volume change, even if other variables remain constant.
Overall, the ideal gas law is a versatile tool for understanding and predicting the behavior of gases under different conditions. It can be used in a wide range of applications, from analyzing the performance of engines to designing industrial processes.
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