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TS EAMCET 7 May 2018 Shift 1 Solved Paper
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© examsnet.com
Question : 127
Total: 160
Kinetic energy in kJ of 280 g of
N
2
at 27°C is approximately (R = 8.314 J
m
o
l
−
1
K
−
1
)
18.7
37.4
56.1
74.8
Validate
Solution:
The number of moles of nitrogen is calculated as:
Moles of
N
2
=
280
28
=
10
The kinetic energy can be calculated as:
K
.
E
.
=
3
n
R
T
2
=
3
×
10
×
8.314
×
300
2
=
37.4
×
10
3
k
J
=
37.4
J
© examsnet.com
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