Certainly! Let's determine the volume of a 0.235-mol ideal gas sample at 1.10 atm and [tex]\(25^{\circ}C\)[/tex].
We will use the Ideal Gas Law for this calculation, which is given by:
[tex]\[ PV = nRT \][/tex]
where
- [tex]\(P\)[/tex] is the pressure,
- [tex]\(V\)[/tex] is the volume,
- [tex]\(n\)[/tex] is the number of moles,
- [tex]\(R\)[/tex] is the ideal gas constant,
- [tex]\(T\)[/tex] is the temperature in Kelvin.
Here are the values provided in the problem:
- [tex]\(n = 0.235 \ \text{mol}\)[/tex]
- [tex]\(P = 1.10 \ \text{atm}\)[/tex]
- [tex]\(T = 25^{\circ}C\)[/tex]
First, we need to convert the temperature from Celsius to Kelvin. The conversion formula is:
[tex]\[ T(K) = T(^{\circ}C) + 273.15 \][/tex]
So, for [tex]\(25^{\circ}C\)[/tex]:
[tex]\[ T = 25 + 273.15 = 298.15 \ \text{K} \][/tex]
The ideal gas constant [tex]\(R\)[/tex] is:
[tex]\[ R = 0.0821 \ \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol}) \][/tex]
Now, we can rearrange the Ideal Gas Law to solve for [tex]\(V\)[/tex] (volume):
[tex]\[ V = \frac{nRT}{P} \][/tex]
Substituting in the given values:
[tex]\[ V = \frac{(0.235 \ \text{mol}) \times (0.0821 \ \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol})) \times (298.15 \ \text{K})}{1.10 \ \text{atm}} \][/tex]
Calculating the numerator:
[tex]\[ (0.235 \ \text{mol}) \times (0.0821 \ \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol})) \times (298.15 \ \text{K}) \approx 5.752787925 \ \text{L} \cdot \text{atm} \][/tex]
Now, divide by the pressure:
[tex]\[ V = \frac{5.752787925 \ \text{L} \cdot \text{atm}}{1.10 \ \text{atm}} \approx 5.2294154772727275 \ \text{L} \][/tex]
Therefore, the volume of the gas sample is approximately [tex]\(5.229 \ \text{L} \)[/tex].
