# If the temperature is increased by a factor of 30 at

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greater force, which increases the size (volume) of the container. If the temperature is increased by a factor of 3.0 (at constant pressure) then the volume will increase by a factor of 3.0(Charles’s law). 1 12211221 11(1)()(1)(3)= = (1)(1)()PV n TVTVP n TTV2= 3V1c) As the number of molecules of gas increases at constant pressure and temperature (Pand Tare fixed), the force they exert on the container increases. This results in an increase in the volume of the container. Adding 3 moles of gas to 1 mole increases the number of moles by a factor of 4, thus the volume increases by a factor of 4(Avogadro’s law). 1 12211221 11(1)()(4)(1)= = (1)()(1)PV n TVnVP n TnV2= 4V15-5
5.21 Plan:Use the relationship 1 12212= PVP VTTor 1 1 222 1= PV TVP T. Rand n are fixed. Solution:a) As the pressure on a fixed amount of gas (nis fixed) decreases at constant temperature (Tis fixed), the molecules move farther together, increasing the volume. When the pressure is reduced by a factor of 4, the volume increases by a factor of 4at constant temperature (Boyle’s law). 1 1 21122 11()()(1)= = (1/ 4)(1)PV TPVVP TPV2=4V1 b) As the pressure on a fixed amount of gas (nis fixed) doubles from 101 kPa to 202 kPa at constant temperature, the volume decreases by a factor of ½. As the temperature of a fixed amount of gas (nis fixed) decreases by a factor of ½ (from 310 K to 155 K) at constant pressure, the volume decreases by a factor of ½. The changes in pressure and temperature combine to decrease the volume by a factor of 4.P1= 760 torr = 101 kPa T1= 37°C + 273 = 310 K 1 1 2122 1(101 kPa)()(155 K)= = (202 kPa)(310 K)PV TVVP TV2=14V1c) As the pressure on a fixed amount of gas (nis fixed) decreases at constant temperature (Tis fixed), the molecules move farther together, increasing the volume. When the pressure is reduced by a factor of 2, the volume increases by a factor of 2at constant temperature (Boyle’s law). T2= 32°C + 273 = 305 K P2= 101 kPa = 1 atm 1 1 2122 1(2 atm)()(305 K)= = (1 atm)(305 K)PV TVVP TV2= 2V15.22 Plan:Use the relationship 1 12212= PVP VTTor 1 1 222 1= PV TVP T. Rand n are fixed. Solution:a)The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2(Charles’s law). 1 1 2122 1(1)()(400 K)= = (1)(800 K)PV TVVP TV2= ½ V1b) T1= 250°C + 273 = 523 K T2= 500°C + 273 = 773 K The temperature increases by a factor of 773/523 = 1.48, so the volume is increased by a factor of 1.48(Charles’s law). 1 1 2122 1(1)()(773 K)= = (1)(523 K)PV TVVP TV2= 1.48V1c) The pressure is increased by a factor of 3, so the volume decreases by a factor of 3(Boyle’s law). 1 1 2122 1(2 atm)()(1)= = (6 atm)(1)PV TVVP TV2= V15.23Plan:Use the relationship 1 1221 122= PVP Vn Tn Tor 1 122221 1= PV n TVP n T. Solution:a) Since the number of moles of gas is decreased by a factor of 2, the volume would be decreased by a factor of 2 (Avogadro’s law).