Question

The pressure of 2 moles of ideal gas at 546k having volume 44.8l is

Answer 1

Answer:

I hope understand this solution

Answer 2

Answer:

The pressure of $2$ moles of an ideal gas at $546K$ temperature having volume $44.8l$ is $2.02 \times 10^{-5} Pa$.

Explanation:

Data given,

The number of moles of an ideal gas, n = $2 mol$

The temperature of an ideal gas, T =$546K$

The volume of an ideal gas, V = $44.8L$ = $0.0448 m^{3}$

The pressure of an ideal gas, P =?

As we know,

• From the ideal gas equation, we can calculate the pressure of an ideal gas:
• $PV =nRT$

Here,

• P = Pressure in pascal
• V = Volume in m³
• n = Number of moles in mol
• R = Gas constant = $8.314 m^{3}-Pa/mol-K$
• T = Temperature in Kelvin

After putting all the given values in the equation, we get:

• $P \times 0.0448 = 2\times 8.314 \times546$
• $P = \frac{9078.88}{0.0448}$
• $P = 2.02 \times 10^{-5} Pa$

Hence, the pressure, P = $2.02 \times 10^{-5} Pa$.