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5th?

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Asked by smanishkumar2002 2nd August 2018, 9:23 PM
Answered by Expert
Answer:
Q5(a)
 
Given:
 
temperature T = 27 °C = 300 K
 
The energy of a monoatomic ideal gas is,
 
straight E space equals space 3 over 2 RT

space space space equals 3 over 2 cross times 1.987 cross times 300

straight E space equals 894.15 space cal
 
The energy of a monoatomic gas is 894.15 cal.
 
Q5 (b)
 
Given:
 
Weight of N =3.45 g
 
Atomic weight of Ne = 20.2 
 
Temperature T1 =0 °C  = 273 K
 
                    T2 = 100 °C  = 373 K
 
 
Change in kinetic energy is given by
 
straight K. straight E space equals 3 over 2 straight R open parentheses straight T subscript 2 minus straight T subscript 1 close parentheses

straight K. straight E. space equals 3 over 2 space cross times 1.987 cross times open parentheses 373 minus 273 close parentheses

space space space space space space space space equals 298.05 space cal

This space is space change space in space kinetic space energy space for space 1 space mol space of space gas.

therefore space For space 3.45 space straight g space of space Ne space

Moles space of space Ne space equals space fraction numerator Weight space of space Ne over denominator Molecular space weight space of space Ne end fraction

space space space space space space space space space space space space space space space space space space space equals fraction numerator 3.45 over denominator 20.2 end fraction

Moles space of space Ne space equals space 0.1707 space mol

Kinetic space enegy space for space 0.1707 space mol space of space Ne space equals space 298.05 space cross times 0.1707 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 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 space space space space space equals 50.90 space cal.
 
Answered by Expert 3rd August 2018, 12:04 PM
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