A sample of `0.20` mole of a gas at `44^@C` and `1.5 atm` pressure is cooled to `27^@C` and compressed to `3.0atm`. Calculate `DeltaV`. Suppose the original sample of gas was heated at constant volume until its pressure was `3.0atm` and then cooled at `27^@C`, what would have been `DeltaV`?

A sample of `0.20` mole of a gas at `44^@C` and `1.5 atm` pressure is

1.	A sample of `0.20` mole of a gas at `44^@C` and `1.5 atm` pressure is cooled to `27^@C` and compressed to `3.0atm`. Calculate `DeltaV`. Suppose the original sample of gas was heated at constant volume until its pressure was `3.0atm` and then cooled at `27^@C`, what would have been `DeltaV`?
Answer» `V_(1)=(nRT_(1))/(P_(1))=(0.2xx0.0821xx317)/1.5=3.470 litre` `V_(2)=(nRT_(2))/(P_(2))=(0.2xx0.821xx300)/3.0` `=1.642 litre` `:. DeltaV=V_(2)-V_(1)=1.642-3.470` `=-1.828 litre` In second case the initial and final states at constant volume are same and thus `DeltaV=0`, since volume is a state function.

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A sample of `0.20` mole of a gas at `44^@C` and `1.5 atm` pressure is cooled to `27^@C` and compressed to `3.0atm`. Calculate `DeltaV`. Suppose the original sample of gas was heated at constant volume until its pressure was `3.0atm` and then cooled at `27^@C`, what would have been `DeltaV`?

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