Question

th sidf you. or a s h. No 3 Problem 11. Why is the gas more easily compressed than a solid or liquid? Joules fibrilla Problem 12. The tires on a track are inflated to 76 psi. Convert this value to atmos- pheres. mic van Problem 13. A sample of helium gas has a volume 8.0 L at a pressure of 3.0 atm. What is the volume of the gas at 6.0 atm? (Use the Boyle's law). ATis Problem 14. A volume of 0.50 L of air at 37 °C is expelled from the lungs into the cold surroundings at 0.0 °C. What volume does the expelled air occupy at this temperature? (Use the Charles's law). Ans Problem 15. A tire of bicycle in a cool garage at 12°C (285K) has a pressure 0.54 atm. Calculate the pressure inside the tire after riding the bike at 45 °C (318 K) (Use the Gay- Lussacs's Law). Ans Problem 16. The pressure inside a 1.0L balloon at 25 °C is 745 mm Hg. What is the pressure inside the balloon when it is cooled to -42°C and expands to 2.0 L in volume? (Use the Combined gas law). Ans

Answer 1

11) because in ceas , the Intermolecular distance is very high tham solid and liquid ,so geeses are easily compressed 12) I psi = 0.068 atm P= 76 psi P = 76 x 0.068 TP = 5.168 cetm I Ane V2=? 13 vi= 8 Lit P2 = 6ath P,= 3 cetm. - PIVI = P2 U2 3x8 = 6*V2 = 2 = 4 Lit Ane v,= o.scit T = 37c =) 37+ 273.15 = 310.15k v2 = 3 T2=0°c ) 273.15k 14) [V2 = 0.366 Lit one

Is) Ti= 28k Pi=o.suatm P2=? T2 = 318k B = 0.60 atm Ane 16) v = 1 uit To 25€ - 298 k T2 = -42 (=231k V₂ = 2lit Py = ? p=445 mmHg T2 745x1 = P x2 Iys x 1 298 231 P₂ = 288 75 mming Ave