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`        A sample of an ideal gas (γ = 1.5) is compressed adiabatically from a volume of 150 cm3 to 50 cm3. The initial pressure and the initial temperature are 150 kPa and 300 K. Find (a) the number of moles of the gas in the sample, (b) the molar heat capacity at constant. Volume, (c) the final pressure and temperature, (d) the work done by the gas in the process and (e) the change in internal energy of the gas.`
5 years ago 396 Points
```							Sol. PV = nRT
Given P = 150 KPa = 150 * 10^3 Pa,  V = 150 cm^3 = 150 * 10^-6 m^3, T = 300 k
(a) n = PV/RT = 150 * 10^3 * 150 * 10^-6/8.3 * 300 = 9.036 * 10^-3 = 0.009 moles.
(b) C base P/C baseV = γ ⇒ γR/( γ - 1)C base v = γ [∴ C base P = γR/ γ - 1]
⇒ C base v = R/γ – 1 = 8.3/1.5 – 1 = 8.3/0.5 = 2R = 16.6 J/mole
(c) Given P base 1 = 150 KPa = 150 × 10 base 3 Pa, P base 2 =?
V base 1 = 150 cm^3 = 150 * 10^-6 m^3, γ = 1.5
V basse 2 = 50 cm^3 = 50 * 10^-6 m^3, T base 1 = 300 k, T base 2 = ?
Since the process is adiabatic Hence – P base 1V base 1^γ
⇒ 150 * 10^3 (150 * 10^-6)^γ = P base 2 * (50 * 10^-6) ^γ
⇒ P base 2 = 150 * 10^3 * (150 * 10^-6/50 * 10^-6)^1.5 = 150000 * 3^1.5 = 779.422 * 10^3 Pa = 780 KPa
(d) ∆Q = W + ∆U or W = -∆U [∴ ∆U = 0, in adiabatic]
= - nC base vdT = - 0.009 * 16.6 * (520 - 300) = - 0.009 * 16.6 * 220 = - 32. 8 J = - 33 j
(e) ∆U = nC base vdT = 0.009 * 16.6 * 220 = 33 J

```
5 years ago
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