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What is the temperature of 0.80 mol of a gas stored in a 275 mL cylinder at 175 kPa?

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

A. 4.6 K
B. 7.2 K
C. 61 K
D. 96 K


Sagot :

To determine the temperature of the gas, we can use the Ideal Gas Law, which is stated as:

[tex]\[ PV = nRT \][/tex]

Here:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( V \)[/tex] is the volume,
- [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 in Kelvin.

We are given:
- [tex]\( n = 0.80 \)[/tex] moles,
- [tex]\( V = 275 \)[/tex] mL,
- [tex]\( P = 175 \)[/tex] kPa,
- [tex]\( R = 8.314 \frac{L \cdot kPa}{mol \cdot K} \)[/tex].

First, convert the volume from milliliters to liters since the ideal gas constant [tex]\( R \)[/tex] is given in terms of liters:
[tex]\[ V = 275 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.275 \, \text{L} \][/tex]

Next, we rearrange the Ideal Gas Law to solve for temperature [tex]\( T \)[/tex]:

[tex]\[ T = \frac{PV}{nR} \][/tex]

Substitute the given values into this formula:

[tex]\[ T = \frac{175 \, \text{kPa} \times 0.275 \, \text{L}}{0.80 \, \text{mol} \times 8.314 \, \frac{L \cdot kPa}{mol \cdot K}} \][/tex]

Now, perform the calculations as follows:

[tex]\[ T = \frac{175 \times 0.275}{0.80 \times 8.314} \][/tex]

[tex]\[ T = \frac{48.125}{6.6512} \][/tex]

[tex]\[ T \approx 7.23553644455136 \, \text{K} \][/tex]

Therefore, the temperature of the gas is approximately [tex]\( 7.2 \, \text{K} \)[/tex].

So the correct answer is:

[tex]\[ \boxed{7.2 \, \text{K}} \][/tex]