A) 60 K
B) 75 K
C) 150 K
D) 225 K
Correct Answer: A
Solution :
Here \[{{V}_{1}}=10\,L,\,{{V}_{2}}=2L\] \[{{\text{P}}_{\text{1}}}\text{=}\,\,\text{1}\,\text{atm,}\,{{\text{P}}_{\text{2}}}\text{=}\,\text{1}\,\text{atm}\] \[{{T}_{1}}=300\,K,{{T}_{2}}=?\] \[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[\frac{1\times 10}{300}=\frac{1\times 2}{{{T}_{2}}}\] \[{{T}_{2}}=\frac{300\times 2}{10}=60\,K\]
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