a gas is collected at a pressure of 95 cm of hg and at a temperature of 50 c to what temperature should it be cooled so that it occupies a volume of 80 percent of its original volume when pressure of the gas is 90 cm of hg
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76 cm = 1 atm
95cm = 5/4 atm = 1.25 atm
T = 50⁰C = 323K
cet the no. of mole = 1
PV = nRT
1.25 × v = 1 × 1/12 × 323
v = 323/12 × 1.25 = 21.533
now, new v = 80% of old v
new v = 17.2264
p = 90/76
PV = nRT
= 90/76 × 17.226
= 1 × 1/12 × t
= t( 90 × 17.226 × 12) ÷ 76
= 244.79
=> 244.8 k
hope it helped!! :)
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