Question

a) Calculate the work when 617 grams of solid CO2 sublimes at 221.5K.
B) Calculate the work when 1 mole of solid CO2 sublimes at 221.5K

Answer 1

Answer on Question #53092 - Chemistry - Analytical Chemistry

a) Calculate the work when 617 grams of solid $\mathrm{CO}_{2}$ sublimes at 221.5K.

b) Calculate the work when 1 mole of solid $\mathrm{CO}_{2}$ sublimes at 221.5K

a) $\mathrm{CO}_{2}(\mathrm{s}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})$

617 grams / 44 gram / moles = 14 moles of $\mathrm{CO}_{2}$

Work done = - pressure-volume-amount of moles of $\mathrm{CO}_{2}$ at 221.5K [ this is the change in volume when 14 moles solid $\mathrm{CO}_{2}$ changes to 14 moles $\mathrm{CO}_{2}$ gas at 221.5K ]

1 mole $\mathrm{CO}_{2}$ occupies volume $V_{1} = 22.4\mathrm{L}$ at STP, $P_{1} = 1$ atm and $T_{1} = 273\mathrm{K}$

At $T_{2} = 221.5\mathrm{K}$ pressure remains 1 atm, the gas occupies volume $V_{2}$

$$P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}$$ $$V_{2} = P_{1}V_{1}T_{2}/T_{1}P_{2} = (1.224.221.5)/(273.1) = 18.2\mathrm{L}$$

Work done = 1 atm · 18.2L (system volume change) · 14 moles of $\mathrm{CO}_{2} = 254.8$ L-atm or 25811.24 J

1 L-atm = 101.3 J

Answer: 25811.24 J

b) $\mathrm{CO}_{2}(\mathrm{s}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})$

Work done = - pressure-volume of 1 mole $\mathrm{CO}_{2}$ at $221.5\mathrm{k}$ [ this is the change in volume when 1 mole solid $\mathrm{CO}_{2}$ changes to 1 mole $\mathrm{CO}_{2}$ gas at $221.5\mathrm{k}$ ]

1 mole $\mathrm{CO}_{2}$ occupies volume $V_{1} = 22.4\mathrm{L}$ at STP, $P_{1} = 1$ atm and $T_{1} = 273\mathrm{K}$

At $T_{2} = 221.5\mathrm{K}$ pressure remains 1 atm, the gas occupies volume V2

$$P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}$$ $$V_{2} = P_{1}V_{1}T_{2}/T_{1}P_{2} = (1.224.221.5)/(273.1) = 18.2\mathrm{L}$$

Work done = 1 atm · 18.2L (system volume change) = 18.2 L-atm or 1844 J

1 L-atm = 101.3 J

Answer: 1844 J

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