Question

A sample of gas (y =1.5) is compressed adiabatically from a volume of 1600 cmºto
400 cm". If the initial pressure is 150 kPa, what is the final pressure and how much
work is done on the gas in the process?

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Answer 1

Answer:

PV = nRT Given P = 150 KPa = 150 × 103 Pa, V = 150 cm3 = 150 × 10–6 m3, T = 300 k = 779.422 × 103 Pa ≈ 780 KPa (d) ΔQ = W + ΔU or W = –ΔU [ΔU = 0, in adiabatic] = – nCVdT = – 0.009 × 16.6 × (520 – 300) = – 0.009 × 16.6 × 220 = – 32.8 J ≈ – 33 J (e) ΔU = nCVdT = 0.009 × 16.6 × 220 ≈ 33 JRead more on Sarthaks.com - https://www.sarthaks.com/63697/a-sample-of-an-ideal-gas-1-5-is-compressed-adiabatically-from-a-volume-of-150-cm-3-to-50-cm-3PV = nRT Given P = 150 KPa = 150 × 103 Pa, V = 150 cm3 = 150 × 10–6 m3, T = 300 k = 779.422 × 103 Pa ≈ 780 KPa (d) ΔQ = W + ΔU or W = –ΔU [ΔU = 0, in adiabatic] = – nCVdT = – 0.009 × 16.6 × (520 – 300) = – 0.009 × 16.6 × 220 = – 32.8 J ≈ – 33 J (e) ΔU = nCVdT = 0.009 × 16.6 × 220 ≈ 33 JRead more on Sarthaks.com - https://www.sarthaks.com/63697/a-sample-of-an-ideal-gas-1-5-is-compressed-adiabatically-from-a-volume-of-150-cm-3-to-50-cm-3

Explanation:

PV = nRT Given P = 150 KPa = 150 × 103 Pa, V = 150 cm3 = 150 × 10–6 m3, T = 300 k = 779.422 × 103 Pa ≈ 780 KPa (d) ΔQ = W + ΔU or W = –ΔU [ΔU = 0, in adiabatic] = – nCVdT = – 0.009 × 16.6 × (520 – 300) = – 0.009 × 16.6 × 220 = – 32.8 J ≈ – 33 J (e) ΔU = nCVdT = 0.009 × 16.6 × 220 ≈ 33 JRead more on Sarthaks.com - https://www.sarthaks.com/63697/a-sample-of-an-ideal-gas-1-5-is-compressed-adiabatically-from-a-volume-of-150-cm-3-to-50-cm-3PV = nRT Given P = 150 KPa = 150 × 103 Pa, V = 150 cm3 = 150 × 10–6 m3, T = 300 k = 779.422 × 103 Pa ≈ 780 KPa (d) ΔQ = W + ΔU or W = –ΔU [ΔU = 0, in adiabatic] = – nCVdT = – 0.009 × 16.6 × (520 – 300) = – 0.009 × 16.6 × 220 = – 32.8 J ≈ – 33 J (e) ΔU = nCVdT = 0.009 × 16.6 × 220 ≈ 33 JRead more on Sarthaks.com - https://www.sarthaks.com/63697/a-sample-of-an-ideal-gas-1-5-is-compressed-adiabatically-from-a-volume-of-150-cm-3-to-50-cm-3PV = nRT Given P = 150 KPa = 150 × 103 Pa, V = 150 cm3 = 150 × 10–6 m3, T = 300 k = 779.422 × 103 Pa ≈ 780 KPa (d) ΔQ = W + ΔU or W = –ΔU [ΔU = 0, in adiabatic] = – nCVdT = – 0.009 × 16.6 × (520 – 300) = – 0.009 × 16.6 × 220 = – 32.8 J ≈ – 33 J (e) ΔU = nCVdT = 0.009 × 16.6 × 220 ≈ 33 JRead more on Sarthaks.com - https://www.sarthaks.com/63697/a-sample-of-an-ideal-gas-1-5-is-compressed-adiabatically-from-a-volume-of-150-cm-3-to-50-cm-3PV = nRT Given P = 150 KPa = 150 × 103 Pa, V = 150 cm3 = 150 × 10–6 m3, T = 300 k = 779.422 × 103 Pa ≈ 780 KPa (d) ΔQ = W + ΔU or W = –ΔU [ΔU = 0, in adiabatic] = – nCVdT = – 0.009 × 16.6 × (520 – 300) = – 0.009 × 16.6 × 220 = – 32.8 J ≈ – 33 J (e) ΔU = nCVdT = 0.009 × 16.6 × 220 ≈ 33 JRead more on Sarthaks.com - https://www.sarthaks.com/63697/a-sample-of-an-ideal-gas-1-5-is-compressed-adiabatically-from-a-volume-of-150-cm-3-to-50-cm-3

Answer 2

To Find :

• Final pressure

• Work done by the gas in an adiabatic process is

Solution :

Final pressure

$\large\sf p_1 {V_1}^{ \gamma} = p_2 {V_1}^{ \gamma} \\ \\ \sf p_2 =p_1 \bigg( \frac{V_1}{V_2} { \bigg)}^{ \gamma} \\ \\ \implies \sf 150 \times \bigg( { \frac{1600}{400} \bigg) }^{ \frac{3}{2}} \\ \\\implies \sf 1200 \: kPa$

Work done by the gas in an adiabatic process is

$\sf W = \frac{p_1V_1 - p_2V_2}{ \gamma - 1} \\ \\ \sf \implies W = \frac{(150 \times 1600) - (1200 \times 400)}{1.5 - 1} \\ \\\sf \implies W = \frac{240 - 480}{0.5} \\ \\ \large\sf \implies W = - 480\:J$