500 mL of nitrogen at 27° C is cooled to -5° C at the same pressure. The new volume becomes

Question

500 mL of nitrogen at 27° C is cooled to -5° C at the same pressure. The new volume becomes (A) 326.32 mL (B) 446.66 mL (C) 546.32 mL (D) 771.56mL

Tardigrade Admin · Accepted Answer

Initial volume, V1 = 500 mL Initial temperature, T1 = 27° C = 27 + 273 = 300 K Final temperature, T2 = - 5 + 273 = 268 K V2=? (V1/T1)=(V2/T2) V2=(V1 T2/T1)=(500×268/300) =446.66 mL

Q. $500 m L$ of nitrogen at $2 7^{\circ} C$ is cooled to $- 5^{\circ} C$ at the same pressure. The new volume becomes

326.32 mL

446.66 mL

546.32 mL

771.56mL

Initial volume, $V_{1} = 500 m L$
Initial temperature,
$T_{1} = 2 7^{\circ} C = 27 + 273 = 300 K$
Final temperature,
$T_{2} = - 5 + 273 = 268 K$
$V_{2} = ?$
$\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$
$V_{2} = \frac{V _{1} T _{2}}{T _{1}} = \frac{500 \times 268}{300}$
$= 446.66 m L$

Q. 500mL of nitrogen at 27∘C is cooled to −5∘C at the same pressure. The new volume becomes