Question

The volume of 1 mol of oxygen gas 22.4 lit at 1 atm pressure calculate the temperature of the gas (R= 0.082 l atm m-¹ K-¹) ​

Answer 1

Answer:

The Temperature of oxygen gas is 273K.

Explanation:

The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:

PV = nRT

Where,  

R is the universal gas constant = $0.082 l atmm^{-1} K^{-1}$  

n = Number of moles = 1  

P = Standard pressure = 1 atm  

V = Volume of the gas = 22.4 lit

To find :

The Temperature of Oxygen gas.

• $T=\frac{PV}{nR}$
• $T=\frac{(1)(22.4)}{1(0.082)}$
• $T=\frac{22.4}{0.082}$

We get T = 273K

Therefore, The Temperature of oxygen gas is 273K.

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