Question

What is the temperature (°C) of 1.75 g of O2 gas occupying 4.60 L at 1.00 atm?

Answer 1

Given- m= 1.75 g  

Now moles of O2 gas= Mass/ Molar mass  

n = 1.75/32  

V= 2L  

P= 1atm  

R=0.0821.....Constant  

As PV=nRT  

=> T=PV/nR  

= 1* 2/ (1.75/32)*0.0821 K  

= 2*32/1.75* 0.0821 K  

= 445.45 K  

= (445.45- 273.15) degrees celsius  

= 172.30 degrees celsius..... ANSWER