Thermodynamics: Real-World Scenarios (4)

easy 2 min read

Question

A pressure cooker contains 2 moles of an ideal diatomic gas at 300K300\,\text{K}. The gas is heated isobarically (constant pressure) until its volume doubles. Find the work done by the gas, the heat absorbed, and the change in internal energy. Take R=8.314J/mol KR = 8.314\,\text{J/mol K}.

Solution — Step by Step

For an isobaric process on an ideal gas, V/T=constantV/T = \text{constant}. Volume doubles, so temperature also doubles:

T2=2T1=600KT_2 = 2T_1 = 600\,\text{K}

So ΔT=300K\Delta T = 300\,\text{K}.

For isobaric processes, W=PΔV=nRΔTW = P\Delta V = nR\Delta T:

W=2×8.314×300=4988.4J4988JW = 2 \times 8.314 \times 300 = 4988.4\,\text{J} \approx 4988\,\text{J}

Positive, because the gas expands and pushes the piston outward.

For a diatomic ideal gas, CV=52RC_V = \frac{5}{2}R. Internal energy depends only on temperature:

ΔU=nCVΔT=2×52×8.314×300=12471J\Delta U = nC_V\Delta T = 2 \times \frac{5}{2} \times 8.314 \times 300 = 12471\,\text{J}

Q=ΔU+W=12471+4988=17459JQ = \Delta U + W = 12471 + 4988 = 17459\,\text{J}

Or equivalently Q=nCPΔTQ = nC_P\Delta T with CP=72RC_P = \frac{7}{2}R, giving 2×72×8.314×300=174592 \times \frac{7}{2} \times 8.314 \times 300 = 17459 J. Cross-check passes.

Final: W4988W \approx 4988 J, ΔU12471\Delta U \approx 12471 J, Q17459Q \approx 17459 J.

Why This Works

Notice how heat splits cleanly: about 71% goes into raising internal energy (heating the gas) and 29% goes into pushing the piston. That ratio depends on CV/CPC_V/C_P, which depends on the gas type — diatomic, monatomic, or polyatomic.

The pressure cooker analogy works because once the regulator opens, the process is approximately isobaric at the design pressure. Steam takes in heat partly to vaporise more water (latent heat, ignored here) and partly to push the regulator weight up (work).

Alternative Method

Using γ=7/5\gamma = 7/5 for diatomic gas:

Q=nCPΔT=γγ1nRΔT=7/52/5×4988=17459JQ = nC_P\Delta T = \frac{\gamma}{\gamma - 1}nR\Delta T = \frac{7/5}{2/5}\times 4988 = 17459\,\text{J}

Same answer, fewer calculations if you remember γ\gamma.

Common Mistake

Students use CVC_V to compute heat in an isobaric process. Wrong — at constant pressure, you must use CPC_P. The relation Q=nCVΔTQ = nC_V\Delta T only applies when volume is constant.

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