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Question 5 (1 point):

What is the temperature in °C of 0.742 moles of gas at 2.09 atm and that occupies 9.87 L?


Sagot :

To determine the temperature of the gas in degrees Celsius given the following values:

1. Amount of gas (moles) = 0.742 moles
2. Pressure = 2.09 atm
3. Volume = 9.87 L
4. Universal gas constant (R) = 0.0821 L·atm/(mol·K)

We can use the ideal gas law, which is stated as:

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

where:
- [tex]\( P \)[/tex] is the pressure of the gas (in atm)
- [tex]\( V \)[/tex] is the volume of the gas (in liters)
- [tex]\( n \)[/tex] is the amount of gas (in moles)
- [tex]\( R \)[/tex] is the universal gas constant (0.0821 L·atm/(mol·K))
- [tex]\( T \)[/tex] is the temperature of the gas (in Kelvin)

First, we need to solve for the temperature [tex]\( T \)[/tex]. Rearrange the ideal gas law to solve for [tex]\( T \)[/tex]:

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

Substitute the given values into the equation:

[tex]\[ T = \frac{(2.09 \, \text{atm}) \times (9.87 \, \text{L})}{(0.742 \, \text{moles}) \times (0.0821 \, \text{L·atm/(mol·K)})} \][/tex]

After performing the calculations, we find:

[tex]\[ T = 338.6229402707236 \, \text{K} \][/tex]

This is the temperature in Kelvin. To convert the temperature from Kelvin to degrees Celsius, use the following relation:

[tex]\[ T(°C) = T(K) - 273.15 \][/tex]

So, we need to subtract 273.15 from the Kelvin temperature to get the Celsius temperature:

[tex]\[ T(°C) = 338.6229402707236 \, \text{K} - 273.15 \, \text{K} \][/tex]

[tex]\[ T(°C) = 65.4729402707236 \, \text{°C} \][/tex]

So, the temperature of the gas is approximately 65.47°C.