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Question
0.040 g of He is kept in a closed container initially at 100.0°C. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J.
Use R = 8.3 J K-1 mol-1
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Solution
Here ,
m = 0.040g
M = 4g
n = \[\frac{0.040}{4} = 0.01 \]
T1 = (100 + 273) K = 373 K
He is a monoatomic gas. Thus,
\[\ C_v = 3 × (\frac{1}{2}R) \]
⇒ Cv = 1.5 × 8.3 = 12.45
Let the initial internal energy be U1 .
Let the final internal energy be U2 .
U2 -U1 = n Cv ( T2 -T1 )
⇒ 0.01 × 12.45( T2 - 373) = 12
⇒ T2 = 469 K
The temperature in °C can be obtained as follows: 469 - 273 = 196° C
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