17 l
25 ml. The volume would not change. Now pressure on the other hand...
468ml
The volume (at 20 degrees celsius and a pressure of 1 atmosphere) is 107.7mL. Both changes in temperature or pressure will change the answer.
Vf = 3.0/75 (150) = 6
Zero. PV = nRT. T = 0, so nRT = 0, and thus PV must be zero also. Since we know the volume is not zero, the pressure must be zero.
A sample of Ar gas occupies a volume of 1.2 L at 125°C and a pressure of 1.0 atm. Determine the temperature, in degrees Celsius, at which the volume of the gas would be 1.0 L at the same pressure.
Using the ideal gas law, we can calculate the final temperature of the Xenon gas. Since the volume remains constant, we can use the combined gas law (P₁/T₁ = P₂/T₂) to solve for the final temperature. Rearranging the equation gives T₂ = (P₂ / P₁) * T₁. Plugging in the values, we get T₂ = (0.100 / 0.570) * 20 degrees Celsius = 3.51 degrees Celsius.
25 ml. The volume would not change. Now pressure on the other hand...
468ml
Using the ideal gas law, V = (nRT)/P where n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, and P is the pressure. Plug in the values: V = (24.7 mol * 0.0821 Latm/molK * 305 K) / 999 ATM = 6.45 L.
Decrease its pressure.
Water is denser than air, meaning there are more water molecules in a given volume than air molecules. This greater density of water results in more molecules colliding with the surface, creating higher pressure. Additionally, water is less compressible than air, so changes in volume have a more pronounced impact on pressure.
The volume (at 20 degrees celsius and a pressure of 1 atmosphere) is 107.7mL. Both changes in temperature or pressure will change the answer.
To find the moles of carbon dioxide, we need to use the ideal gas law equation: PV = nRT. First, convert the temperature to Kelvin by adding 273.15 (50°C + 273.15 = 323.15 K). Then, plug in the values for P, V, R, and T to solve for n (moles). After calculations, you will find the number of moles of carbon dioxide in the given sample.
the 30L water vapor sample is a gas gases exert pressure on their surroundings thus taking on the shape of their containers the gas is under 1 ATM of pressure so it exerts pressure in return and the volume becomes 30L when the presures are equal if the temperature were to decrease to room temperature, the water vapor would condense to a liquid and the volume would decrease liquids have a volume but no definite shape while gases have neither a definite volume or shape
Using the ideal gas law, V1/T1 = V2/T2, where V is volume and T is temperature, we can set up an equation to solve for the final temperature. Plugging in the given values, we get 49mL/7°C = 74mL/T2. Solving for T2 gives approximately 10.9°C.
Vf = 3.0/75 (150) = 6