If the gas was instead in a cylinder with a floating, massless, frictionless piston, what would the volume of the gas be (in liters) at STP?
You need to post ALL of the problem.
V1P1/T1 = V2P2/T2 Temp in Kelvin and assuming room temp of 25C(298K)
7.5×150/293 = V2x1/298
V2 = 1144L
Remeber to round your answer to 2 sig figs:
1.0 x 10^3
To find the volume of the gas in a cylinder with a floating, massless, frictionless piston at STP (Standard Temperature and Pressure), we need to understand the conditions for STP.
At STP, the temperature is 0 degrees Celsius (273.15 Kelvin) and the pressure is 1 atmosphere (atm).
The volume of the gas can be calculated using the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = universal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
Since the gas is at STP, we can assume that one mole of the gas is present. Hence, n = 1 mole.
Substituting the values into the equation, we have:
(1 atm)(V) = (1 mole)(0.0821 L·atm/mol·K)(273.15 K)
Simplifying the equation, we get:
V = (0.0821 L·atm/mol·K)(273.15 K)
Calculating the value, we have:
V ≈ 22.4 L
Therefore, the volume of the gas in a cylinder with a floating, massless, frictionless piston at STP is approximately 22.4 liters.
To determine the volume of the gas in the given scenario (floating, massless, frictionless piston in a cylinder) at STP (Standard Temperature and Pressure), we need to consider the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
At STP, the temperature is 273.15 Kelvin, and the pressure is 1 atm.
Considering that the piston is floating and there is no external force acting on it, it means that the pressure inside the cylinder is equal to the atmospheric pressure. Therefore, P = 1 atm.
Let's assume we have one mole of gas (n = 1).
Using the ideal gas law equation, we can rearrange it to calculate the volume:
V = nRT / P
V = (1 mol) * (0.0821 L.atm/(mol.K)) * (273.15 K) / (1 atm)
V = 22.41 L
Therefore, the volume of the gas in the cylinder at STP would be approximately 22.41 liters.