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A `10.0 L` container is filled with a gas to a pressure of `2.00 atm` at `0^(@)C`. At what temperature will the pressure inside the container be `2.50 atm` ? |
Answer» As the volume of the container remains constant, applying pressure-temperature law,viz. `(P_(1))/(T_(1))=(P_(2))/(T_(2))` We get `(2 atm)/(273 K)=(2.50 atm)/(T_(2))` or `T_(2)=341 K=341-273^(@)C=68^(@)C` |
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