A $2.00 m o l$ sample of helium gas initially at $300 K$ and $0.400 a t m$ is compressed isothermally to $1.20 a t m$ . Noting that the helium behaves as an ideal gas, find
the final volume of the gas,
the work done on the gas and
the energy transferred by heat.

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Solution

The final volume will be one-third of the original, for the temperature to be constant, the sample must liberate as many joules by heat as it takes in by work.
The ideal gas law can tell us the original and final volumes. The negative integral of $p d V$ will tell us the work input, and the first law of thermodynamics will tell us the energy output by heat.
1. Rearranging $P V = n R T$ , we get $V_{i} = \frac{n R T}{P_{i}}$
The initial volume is
$V_{i} = \frac{(2.00 m o l) (8.314 J / m o l K) (300 K)}{(0.400 a t m) (1.013 \times 10^{5} P a / a t m)} (\frac{1 P a}{N / m^{2}}) = 0.123 m^{3}$
For isothermal compression, $P V$ is constant, so $P_{i} V_{i} = P_{f} V_{f}$ and the final volume is
$V_{f} = V_{i} \frac{P_{i}}{P_{f}} = (0.123 m^{3}) (\frac{0.400 a t m}{1.20 a t m}) = 0.0410 m^{3}$ .
2. $W = - \int p d V = - \int \frac{n R T}{V} d V = - n R T ln (\frac{V_{f}}{V_{i}})$ .
$= - (4988 J) ln (\frac{1}{3}) = + 5.48 k J$
3. The ideal gas keeps constant temperature so $Δ E_{i n t} = 0 = Q + W$ and the heat is $Q = - 5.48 k J$ .

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A 2.00mol sample of helium gas initially at 300K and 0.400atm is compressed isothermally to 1.20atm. Noting that the helium behaves as an ideal gas, findthe final volume of the gas,the work done on the gas andthe energy transferred by heat.

A $2.00 m o l$ sample of helium gas initially at $300 K$ and $0.400 a t m$ is compressed isothermally to $1.20 a t m$ . Noting that the helium behaves as an ideal gas, find
the final volume of the gas,
the work done on the gas and
the energy transferred by heat.