what is the approximate volume of gas in a 1.50 mol sample that exerts a pressure of 0.922 atm and has a temp of 10.0°C?
To find the approximate volume of gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature of the gas (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 10.0 + 273.15
T(K) = 283.15 K
Now we can substitute the values into the ideal gas law equation:
(0.922 atm) * V = (1.50 mol) * (0.0821 L·atm/(mol·K)) * (283.15 K)
Simplifying the equation:
0.922 V = 0.124535175
Dividing both sides of the equation by 0.922:
V ≈ 0.135 L
Therefore, the approximate volume of the gas is 0.135 liters.
To find the approximate volume of gas, you can use the ideal gas law, which states that 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.
To solve for the volume, we need to rearrange the equation as V = (nRT) / P. Let's go step by step to calculate it:
1. Convert the temperature from Celsius to Kelvin. Kelvin is an absolute temperature scale, so we need to add 273.15 to the Celsius temperature. In this case, 10.0°C + 273.15 = 283.15 K.
2. Plug in the given values into the equation:
n = 1.50 mol (number of moles)
R = 0.0821 atm L / (mol K) (ideal gas constant)
T = 283.15 K (temperature in Kelvin)
P = 0.922 atm (pressure)
V = (1.50 mol * 0.0821 atm L / (mol K) * 283.15 K) / 0.922 atm
3. Calculate the volume:
V = 35.14 L
Therefore, the approximate volume of gas in the 1.50 mol sample is 35.14 liters.