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प्रश्न
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.
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उत्तर
The ideal gas equation is
PV = nRT
Given, P1 = 150 kPa = 150 × 103 Pa
V1 = 150 cm3 = 150 × 10−6 m3
T1= 300 K
(a)
`"n" =("P""V")/("R""T") = 9.036 xx 10^-3`
n = 0.009
(b)
`"C"_"p"/"C"_"v" = gamma , "C"_"p" -"C"_"v" ="R"`
So, `"C"_"v" = "R"/(gamma-1) = 2"R" = 8.3/0.5 = 16.6 "J" "mol" -"K"`
(c) Given,
P1 = 150 kPa = 150 × 103 Pa
P2 = ? V1 = 150 cm3
= 150 × 10−6 m3
γ = 1.5
V2 = 50 cm3 = 50 × 10−6 m3,
T1 = 300 K
T2 = ?
Since the process is adiabatic, using the equation of an adiabatic process,we get
P1V1γ = P2V2γ
⇒ 150 × 103 × (150 × 10−6)γ = P2 × (50 × 10−6)γ
`=> "P"_2 = 150 xx 10^3 xx ((150 xx 10^-6)^1.5)/(50 xx 10 ^-6)^1.5`
P2 = 150000 × (3)1.5
P2 = 779.422 × 103 Pa
P2 = 780 kPa
Again,
P11−γ T1γ = P11−γ T2γ
⇒ (150 × 103)1−1.5 × (330)1.5 = (780 × 103)1−1.5 × T21.5
⇒ T21.5 = (150 × 103)1−1.5 × (300)1.5 × 3001.5
T21.5 = 11849.050
⇒ T2 = (11849.050)1/1.5
T2 = 519.74 = 520 K
d) dQ = dW + dU
Or dW = −dU [ Since dQ = 0 in an adiabatic process]
dW = −nCvdT
dW = −0.009 × 16.6 × (520 − 300)
dW = −0.009 × 16.6 × 220
dW = −32.87 J ≈ −33 J
(e)
dU = nCvdT
dU = 0.009 × 16.6 × 220 ≈ 33 J
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