A glass bulb of volume 400 cm3 is connected to another of volume 200 cm3 by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 20.0C and 1.0 atm. The larger bulb is immersed in steam at 100 0C; the smaller, in melting ice at 0 0C. Find the final common pressure.
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V1 = 400 cm³ T = 293°K P = 1 atm n1 = 2 x
V2 = 200 cm³ T = 293°K P = 1 atm n2 = x
Total moles of the gas = 3 x units
P V1 = n1 R T => x = 200/(293 R) units
After changing the temperatures:
P' * 400 = n R 373 and P' 200 = (600 /293R - n) * R 273
Taking their ratio:
=> 2 (600/293R - n) 273 = n 373
=> n = 1200* 273 /(293 * 919 * R) units
P' = 1200 * 373 * 273 /(293 * 919 * 400) = 1.134 atm
V2 = 200 cm³ T = 293°K P = 1 atm n2 = x
Total moles of the gas = 3 x units
P V1 = n1 R T => x = 200/(293 R) units
After changing the temperatures:
P' * 400 = n R 373 and P' 200 = (600 /293R - n) * R 273
Taking their ratio:
=> 2 (600/293R - n) 273 = n 373
=> n = 1200* 273 /(293 * 919 * R) units
P' = 1200 * 373 * 273 /(293 * 919 * 400) = 1.134 atm
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