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Q) An open bulb containing air at 19 degree Celsius was cooled to a certain temperature at which the number of moles of gaseous molecules increased by 25% what is the final temperature?
Asked by Anish | 08 Jan, 2019, 11:22: PM
answered-by-expert Expert Answer
Let,
 
Pressure in bulb = P1 
 
Volume of air = V
 
Initial temperature = T1 = 19+273 
 
                                    =292 K
 
Final temperature = T2 
 
Initial moles = n1 
 
Final moles = n2 
 
We have,
 
n2 = 1.25n1 
 
Using ideal gas equation,
 
 
PV equals space nRT Initially comma PV equals space straight n subscript 1 RT subscript 1 Finally comma PV equals space straight n subscript 2 RT subscript 2 therefore space straight n subscript 1 straight T subscript 1 space end subscript space equals space straight n subscript 2 straight T subscript 2 straight T subscript 2 space space end subscript space equals space fraction numerator space straight n subscript 1 straight T subscript 1 space end subscript space over denominator space straight n subscript 2 end fraction space space space space space space equals space fraction numerator up diagonal strike space straight n subscript 1 end strike cross times 292 over denominator 1.25 space space up diagonal strike straight n subscript 1 end strike end fraction straight T subscript 2 space space end subscript space equals space 233.6 space straight K
 
T2 = 233.6 K = 233.6 − 273
 
     = −39.6 °C
 
 
The final temperature is −39.6 °C
 
Answered by Varsha | 09 Jan, 2019, 03:06: PM

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Q) An open bulb containing air at 19 degree Celsius was cooled to a certain temperature at which the number of moles of gaseous molecules increased by 25% what is the final temperature?
Asked by Anish | 08 Jan, 2019, 11:22: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
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