A) 2.5 g
B) 2.0 g
C) 1.5 g
D) 1.0 g
Correct Answer: D
Solution :
[d] \[PV=mrT\] Since P, V, r ® remains same |
Hence \[m\propto \frac{1}{T}\]Þ \[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\]Þ \[\frac{13}{{{m}_{2}}}=\frac{(273+52)}{(273+27)}=\frac{325}{300}\] |
Þ m2 = 12 gm |
i.e., mass released = 13gm - 12gm = 1gm. |
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