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

Initial volume, V1=500 mL Initial temperature, T1=27∘C=27+273=300K Final temperature, T2=−5+273=268K V2=? T1V1=T2V2 V2=T1V1T2=300500×268 =446.66 mL

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500 mL of nitrogen at {27}^{{}}{C} is cooled to -{5}^{{}}{C} a

Question

500 mL of nitrogen at $27^{\circ} C$ is cooled to $- 5^{\circ} C$ at the same pressure. The new volume becomes

326.32 mL

446.66 mL

546.32 mL

771.56 mL

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Olivia Discussed

500 mL of nitrogen at

27^{\circ} C

is cooled to

- 5^{\circ} C

at the same pressure. The new volume becomes

15 mins ago

Discuss this question LIVE

15 mins ago

Initial volume,

V_{1} = 500 mL

Initial temperature,

T_{1} = 27^{\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 mL

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Question Text	500 mL of nitrogen at $27^{\circ} C$ is cooled to $- 5^{\circ} C$ at the same pressure. The new volume becomes
Answer Type	Text solution:1
Upvotes	150