What is the pressure of 0.540 mol of an ideal gas at 35.5 L and 223 K?

Use [tex] PV = nRT [/tex] and [tex] R = 8.314 \frac{ L \cdot kPa }{ mol \cdot K } [/tex].

A. 0.715 kPa
B. 2.45 kPa
C. 28.2 kPa
D. 62.7 kPa



Answer :

Certainly! Let's solve this problem step-by-step using the ideal gas law equation:

The ideal gas law is given by:
[tex]\[ P V = n R T \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure of the gas.
- [tex]\( V \)[/tex] is the volume of the gas.
- [tex]\( n \)[/tex] is the number of moles of the gas.
- [tex]\( R \)[/tex] is the ideal gas constant.
- [tex]\( T \)[/tex] is the temperature of the gas in Kelvin.

Here are the given values:
- [tex]\( n = 0.540 \)[/tex] moles
- [tex]\( V = 35.5 \)[/tex] liters
- [tex]\( T = 223 \)[/tex] K
- The ideal gas constant, [tex]\( R = 8.314 \frac{L \cdot kPa}{mol \cdot K} \)[/tex]

We need to find the pressure [tex]\( P \)[/tex].

Rearrange the ideal gas law equation to solve for [tex]\( P \)[/tex]:
[tex]\[ P = \frac{n R T}{V} \][/tex]

Substitute the given values into the equation:
[tex]\[ P = \frac{0.540 \times 8.314 \times 223}{35.5} \][/tex]

Calculate the numerator:
[tex]\[ 0.540 \times 8.314 \times 223 = 1003.24892 \][/tex]

Calculate the pressure [tex]\( P \)[/tex]:
[tex]\[ P = \frac{1003.24892}{35.5} \approx 28.202024788732395 \, kPa \][/tex]

Rounded to one decimal place, the pressure is:
[tex]\[ P \approx 28.2 \, kPa \][/tex]

Thus, the pressure of the gas is:
[tex]\[ \boxed{28.2 \, kPa} \][/tex]