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At what temperature (in Kelvin) will 0.011 mol of argon gas have a volume of 367 mL at 8.48 atm?

*Round your answer to the nearest whole number. Do NOT include units in your answer.*

User Shorena
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1 Answer

4 votes

To solve this problem, you can use the Ideal Gas Law, which is represented by the equation:

PV = nRT

where:

- P is the pressure (in atmospheres),

- V is the volume (in liters),

- n is the number of moles,

- R is the ideal gas constant (0.0821 L.atm/(mol·K)),

- T is the temperature (in Kelvin).

First, you need to convert the volume from milliliters to liters, and the pressure from atm to Pa (1 atm = 101325 Pa).

V = 367 mL / 1000 L

P = 8.48 atm * 101325 Pa/atm

Now, you can rearrange the Ideal Gas Law to solve for temperature:

T = PV / nR

Substitute the known values:

T = (8.48 * 101325) * (367/1000) / (0.011 * 0.0821)

Now, calculate the temperature in Kelvin and round to the nearest whole number.

T ≈ (859468 * 0.367) / 0.0009021

T ≈ 315075.356 / 0.0009021

T ≈ 349745966.61

Now, round to the nearest whole number:

T ≈ 349745967

So, at a temperature of approximately 349,745,967 K, 0.011 mol of argon gas will have a volume of 367 mL at 8.48 atm.

User Ahmed Al Hafoudh
by
8.3k points

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