A sample of N2 has a volume of 616 mL at 68 °C. What will be its volume at -17 °C?

462.5ml

Explanation:

Using Charle's law equation;

V1/T1 = V2/T2

Where;

V1 = initial volume (ml)

V2 = final volume (ml)

T1 = initial temperature (K)

T2 = final temperature (K)

Based on the information provided in this question;

V1 = 616ml = 616/1000 = 0.616L

V2 = ?

T1 = 68°C = 68 + 273 = 341K

T2 = 17 °C = -17 + 273 = 256K

V1/T1 = V2/T2

0.616/341 = V2/256

Cross multiply

341 × V2 = 256 × 0.616

341V2 = 157.69

V2 = 157.69/341

V2 = 0.4625 L or 462.5ml

a is the answer to this question

explanation: