What is the pressure in atm exerted by 2.48 moles of a gas in a 250.0 mL container at 58 degrees celsius

1 answer:

7 0

Since the question manages to include moles, pressure, volume, and temperature, then it is evident that in order to find the answer we will have to use the Ideal Gas Equation: PV = nRT (where P = pressure; V = volume; n = number of moles; R = the Universal Constant [0.082 L·atm/mol·K]; and temperature.

First, in order to work out the questions, there is a need to convert the volume to Litres and the temperature to Kelvin based on the equation:
250 mL = 0.250 L
58 °C = 331 K

Also, based on the equation P = nRT ÷ V

⇒ P = (2.48 mol)(0.082 L · atm/mol · K)(331 K) ÷ 0.250 L
⇒ P = (67.31 L · atm) ÷ 0.250 L
⇒ P = 269.25 atm

Thus the pressure exerted by the gas in the container is 269.25 atm.

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T_1=27°C
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Using Charles law

$\boxed{\sf \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}$

$\\ \sf\longmapsto V_1T_2=V_2T_1$

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