Explanation Explaination: Combined gas law is nothing but combination of three gas laws for studying the effect of variables on the third variable at constant number of moles of gas P 1 V 1 T 1 = P 2 V 2 T 2 Here P 1 is initial pressure, V1 is initial volume, T1 is initial temperature, P 2 is final pressure, V2 is final volume, and T2 is final temperature, Given: After converting temperature in kelvin and pressure in atm: (a)T 1 =298 K T 2 =325 K P 1 =1.12atm P 2 =? V 1 =6. 50 L V 2 = 1.850L (b)T 1 =298 K T 2 =285 K P 1 =1.120 atm P 2 =? V 1 = 6.50 L V 2 = 2.25 L (c) T 1 =298 K T 2 =320 K P 1 =1.12 atm P 2 =? V 1 = 6.50 L V 2 = 12.8 L Formula used: Using combine gas law, P 1 V 1 T 1 = P 2 V 2 T 2 P 2 = P 1 1 × V 1 V 2 × T 2 T 1 Calculation: (a) We must change the temperature in to kelvin T 1 =25 o C+273=298 K T 2 =325 K We change the initial pressure of mm Hg to atm by equality and coversion factors 1 atm=760 mm Hg The conversion factor is given below 1 a t m 760 m m H g o r 760 m m H g 1 a t m P 1 =848 m m H g × 1 a t m 760 m m H g =1.12atm T 1 =298 K T 2 =325 K P 1 =1.12atm P 2 =? V 1 =6. 50L V 2 = 1.850L Using combine gas law, we can solve for P 2 P 1 V 1 T 1 = P 2 V 2 T 2 P 2 = P 1 1 × V 1 V 2 × T 2 T 1 Substitute the values into the gas law to solve P 2 = 1.12 a t m 1 × 6.50 L 1.850 L × 325 K 298 K =4

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Chapter 8.5, Problem 8.33QAP

Interpretation Introduction

To determine: Final pressure of gas in atm.

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Chapter 8.5, Problem 8.34QAP

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