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 \[\to \] remains same |
| Hence, |
| \[m\,\infty \frac{1}{T}\Rightarrow \frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\Rightarrow \frac{13}{{{m}_{2}}}=\frac{\left( 273+52 \right)}{\left( 273+27 \right)}=\frac{325}{300}\] |
| \[\Rightarrow {{m}_{2}}=12gm\] |
| i.e., mass released \[=13\text{ }gm-12gm=1\text{ }gm\] |
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