A) 32.76
B) 23.76
C) 12.05
D) 11.56
Correct Answer: A
Solution :
Given conditions Final conditions \[{{V}_{1}}=38.\text{ }0\text{ }mL,\] \[{{V}_{2}}=?\] \[{{p}_{1}}=746.5-26.\text{ }5\text{ }mm,\] \[{{p}_{2}}=760mm\] \[=720\text{ }mm\] \[{{T}_{1}}=27+273=300\text{ }K\] \[{{T}_{2}}=0+273=273K\] By applying the general gas equation \[\frac{760\times {{V}_{2}}}{273}=\frac{720\times 38}{300}\] \[\therefore \] \[{{V}_{2}}=\frac{720\times 38}{300}\times \frac{273}{760}\] \[=32.76\text{ }mL\] Volume of nitrogen at\[0{}^\circ C\]and 760 mm pressure \[=32.76\text{ }mL\]
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