Question

If 45 g of O2 gas in a 500 mL container is exerting a pressure of 5.2 atm, what temperature is the gas?

Answer 1

If 45 g of O2 gas in a 500 mL container is exerting a pressure of 5.2 atm, 22.53K is the temperature in the gas.

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 5.2 atm

V= 500 mL =0.5 L

n=?

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$

T=?

$Moles = \frac{mass}{molar \;mass}$

$Moles = \frac{45 g}{32}$

$Moles = \frac{45 g}{32}$

Moles = 1.40625

Putting value in the given equation:

$\frac{PV}{RT}=n$

$1.40625= \frac{5.2 \;atm\; X \;0.5 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X T}$

T= 22.53167034 K= 22.53K

Hence, If 45 g of O2 gas in a 500 mL container is exerting a pressure of 5.2 atm, 22.53K is the temperature in the gas.

Learn more about the ideal gas here:

brainly.com/question/27691721

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