A sample of helium behaves as an ideal gas as it is heated at constant pressure

Question

A sample of helium behaves as an ideal gas as it is heated at constant pressure from 273 K to 383 K. If 25.0 J of work is done by the gas during this process, what is the mass of helium present?

Explanation / Answer

1) Work by the gas is given by the work integral W = ? p dV from Vi to Vf for constant pressure process W = p ? dV from Vi to Vf = p·(Vf - Vi) = p·Vf - p·Vi use ideal gas law to express the product of pressure and volume in terms of number of moles and temperature p·V = n·R·T => W = n·R·Tf - n·R·Ti = n·R·(Tf - Ti) solve for n n = W / (R·(Tf - Ti) ) multiply by the molar mass to get the mass n = M·n = M·W / (R·(Tf - Ti) ) = 4g/mol · 20J / (8.314472J/molK · (363K - 273K) ) = 0.107g (2) (a) The internal of energy of an ideal gas is only function of temperature: ?U = ? n·Cv dt from Ti to Tf Since temperature is held constant internal energy doesn't change. ?U = 0 (b) Internal energy change equals heat transferred to plus work done by the gas: ?U = Q + W since ?U = 0 Q = -W = -90J Negative sign indicates that heat is transferred from the gas, that means it must be cooled to maintain process temperature of the compression process.

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