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CBSE Class 11-science Answered

5.18 sum plz
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Asked by lovemaan5500 | 26 Jan, 2019, 12:53: PM
answered-by-expert Expert Answer
Note that, on heating a gas in a vessel, it is the moles of gas which go out, the volume of gas remains constant.
 
Let,
 
Initial moles of gas at T1 = 298 K = n
 
On heating 1/5th  of moles of gas escapes out at T2.
 
Moles space of space gas space left space equals space straight n minus space 1 fifth straight n

space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 over 5 straight n

Now comma

straight n subscript 1 straight T subscript 1 space equals space straight n subscript 2 straight T subscript 2

straight n cross times 298 space equals space 4 over 5 straight n cross times straight T subscript 2

straight T subscript 2 space equals space fraction numerator up diagonal strike straight n cross times 298 cross times 5 over denominator 4 up diagonal strike straight n end fraction

space space space space equals space 1490 over 4

straight T subscript 2 space equals 372.5 space straight K

space space space space equals space 372.5 space minus space 273 space

space space space space equals space 99.55 space to the power of degree straight C
 
The required temperature is 99.55 °C.
 
 
 
Answered by Varsha | 28 Jan, 2019, 12:24: PM
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