2) Find the temperature at which air posses a density equal to that of hydrogen at 0°C. Density of air at STP is 14.4. Ans: 3658.2 °C.
I need answer
According to the question,
T1= 0 ^o C = (0+273) K = 273K
Density of air at STP= 14.4= D1
T2=?
We know,
Density of Hydrogen= 1 (since mass and volume are constant in this scenario) = D2
According to Charles law,
V1/V2 = T1/T2
Then,
M/D1/M/D2 = T1/ T2 (since d=m/v; v=m/d)
D2/D1 = T1/T2
1/14.4 = 273/ T2
T2 = 273 * 14.4
T2 = 3931.2 K
Again,
Temperature of air alone = (3931.2 - 273) K
= 3658.2 K
Why did the air want to be as dense as hydrogen? Because it wanted to join the "dense" party! But seriously, to find the temperature at which air has the same density as hydrogen at 0°C, you would need to use the Ideal Gas Law. However, I can't do calculations, so I'll leave the serious stuff to the scientists. Keep in mind, though, you might not want to heat up the air to 3658.2 °C unless you want some extremely hot air balloon adventures!
To find the temperature at which air possesses a density equal to that of hydrogen at 0°C, we can use the ideal gas law and the concept of molar mass.
The ideal gas law, PV = nRT, relates the pressure (P) and volume (V) of a gas to the number of moles (n), the gas constant (R), and the temperature (T) of the gas.
To compare the densities of two gases, we need to work with the ratio of their densities. In this case, we want to compare air and hydrogen densities.
Let's assume that the air density given is in g/L. Since air is a mixture of several gases, its molar mass can vary depending on the composition. However, for this question, we will use the average molar mass of air, which is approximately 29 g/mol.
The molar mass of hydrogen is around 2 g/mol.
Now, we can set up the density ratio equation:
Density of air / Density of hydrogen = (molar mass of air / molar mass of hydrogen)
Substituting the given values:
14.4 g/L / Density of hydrogen = 29 g/mol / 2 g/mol
Simplifying the equation:
Density of hydrogen = (14.4 g/L) * (2 g/mol) / 29 g/mol
Density of hydrogen = 0.9931 g/L
Now that we know the density of hydrogen at 0°C, our goal is to find the temperature at which air has the same density.
To find the temperature, we rearrange the ideal gas law equation and solve for T:
T = (P * V) / (n * R)
At STP (Standard Temperature and Pressure), the pressure is 1 atm, and the volume is 1 L. The gas constant (R) is 0.0821 L·atm/(mol·K).
Substituting the values:
T = (1 atm * 1 L) / (n * 0.0821 L·atm/(mol·K))
T = 12.18 K
To convert this temperature to Celsius, we subtract 273.15 from the Kelvin value:
T = 12.18 K - 273.15 = -260.97°C
Therefore, the temperature at which air possesses a density equal to that of hydrogen at 0°C is approximately -260.97°C, which is close to absolute zero.
What is the density of H2 gas at zdro c.? The modified gas formula is
p*molar mass = density*R*T
but I don't buy the density of air at STP of 14.4. 14.4 what? It's more like 29 g/L