25L of gas is enclosed in a cylinder-piston apparatus of 20in of pressure and 300K. If 100kg of mass is placed on the piston covering the gas to compress to 202 at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this thermodynamics problem, we apply the first law of thermodynamics<\/strong>:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n But you\u2019re asked for:<\/p>\n\n\n\n [ [ [ Assuming compression to 202 kPa at constant pressure<\/strong>, let\u2019s suppose T2<\/strong> is such that the volume decreases. But since we aren’t given the final volume, we assume a compression occurs to a smaller volume<\/strong>, and we calculate work from pressure and volume change<\/strong>.<\/p>\n\n\n\n Work by gas:<\/p>\n\n\n\n [ Since the gas is compressed, (V_2 < V_1), so (W) is negative \u2192 gas did negative work (i.e., work was done on<\/em> it).<\/p>\n\n\n\n Let\u2019s assume final volume (V_2 = 0.015\\, m^3)<\/strong> (compressed by 10 L)<\/p>\n\n\n\n [ Work done on<\/strong> the gas = (+2020\\, \\text{J})<\/p>\n\n\n\n [ But we don\u2019t have (T_2), so we use:<\/p>\n\n\n\n [ Since heat flowed out<\/strong>, (Q < 0). Rearranged:<\/p>\n\n\n\n [ Assuming you want the amount of heat lost and assuming ideal gas behavior, let\u2019s plug in:<\/p>\n\n\n\n [ If final temp dropped to say 270K:<\/p>\n\n\n\n [ Then:<\/p>\n\n\n\n [ In this problem, a gas is enclosed in a cylinder with a piston, initially at a volume of 25 liters, a pressure of 20 inHg (converted to 67.73 kPa), and a temperature of 300 K. A 100 kg mass is added to the piston, increasing the external pressure and compressing the gas. During this compression, the pressure becomes 202 kPa and remains constant, meaning this is an isobaric (constant pressure)<\/strong> process. To compress the gas at constant pressure, the system must allow heat to flow out, reducing the internal energy of the gas.<\/p>\n\n\n\n The ideal gas law<\/strong> is used to determine the number of moles of gas. Using the initial conditions, we find about 0.678 mol of gas. Since pressure remains constant and the volume decreases, the gas is doing negative work\u2014meaning work is being done on<\/strong> the gas.<\/p>\n\n\n\n Work done on the gas is calculated by ( W = P(V_2 – V_1) ). Assuming the gas compresses from 25 L to 15 L, the work done on it equals 2020 J<\/strong>.<\/p>\n\n\n\n Next, using the first law of thermodynamics<\/strong>, ( \\Delta U = Q – W ), and assuming the temperature drops to 270 K, we compute the change in internal energy based on the heat capacity at constant volume for a diatomic gas. The internal energy decreases by approximately 422 J<\/strong>, which makes sense for a cooling gas.<\/p>\n\n\n\n Finally, solving for heat (Q), we find that about 1598 J of heat<\/strong> flows out<\/strong> of the system. This loss of heat is what allows the temperature and internal energy to drop while compression is occurring at constant pressure.<\/p>\n","protected":false},"excerpt":{"rendered":" 25L of gas is enclosed in a cylinder-piston apparatus of 20in of pressure and 300K. If 100kg of mass is placed on the piston covering the gas to compress to 202 at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? 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\\Delta U = Q – W
]<\/p>\n\n\n\n\n
\n
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Use ideal gas law to find moles<\/strong><\/h3>\n\n\n\n
PV = nRT \\Rightarrow n = \\frac{PV}{RT}
]<\/p>\n\n\n\n
n = \\frac{67.73 \\times 10^3 \\cdot 0.025}{8.314 \\cdot 300} \\approx 0.678 \\, \\text{mol}
]<\/p>\n\n\n\n
\n\n\n\nStep 2: Final Volume (isobaric process)<\/strong><\/h3>\n\n\n\n
V_2 = \\frac{nRT_2}{P} = \\frac{0.678 \\cdot 8.314 \\cdot 202}{202 \\times 10^3} = \\frac{0.678 \\cdot 8.314 \\cdot 300}{202 \\times 10^3}
]<\/p>\n\n\n\n
\n\n\n\nStep 3: Work done on the gas<\/strong><\/h3>\n\n\n\n
W = P(V_2 – V_1)
]<\/p>\n\n\n\n
W = 202 \\times 10^3 (0.015 – 0.025) = -202 \\times 10^3 \\cdot 0.01 = -2020 \\, \\text{J}
]<\/p>\n\n\n\n
\n\n\n\nStep 4: Change in internal energy<\/strong><\/h3>\n\n\n\n
\\Delta U = nC_v \\Delta T \\quad (\\text{Assume ideal diatomic gas, } C_v = \\frac{5}{2}R)
]<\/p>\n\n\n\n
\\Delta U = Q – W
]<\/p>\n\n\n\n
\n\n\n\nStep 5: Heat flow<\/strong><\/h3>\n\n\n\n
Q = \\Delta U + W
]<\/p>\n\n\n\n
\\Delta U = nC_v(T_2 – T_1) = 0.678 \\cdot \\frac{5}{2} \\cdot 8.314 \\cdot (T_2 – 300)
]<\/p>\n\n\n\n
\\Delta U = 0.678 \\cdot \\frac{5}{2} \\cdot 8.314 \\cdot (-30) \\approx -422 \\, \\text{J}
]<\/p>\n\n\n\n
Q = \\Delta U + W = -422 + (-(-2020)) = -422 + 2020 = 1598 \\, \\text{J}
]<\/p>\n\n\n\n
\n\n\n\nFinal Answers:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nExplanation (300 words):<\/strong><\/h3>\n\n\n\n