Question

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.

Answer 1

• a) Number of moles of the gas in the sample is 0.009 moles
• b) Molar heat capacity at constant volume is 16.65.
• c) Final pressure and temperature is 780 k Pa and 520 K respectively
• d) Work done by the gas is -33 J.
• e) Change in internal energy of the gas is 33 J.

a) Moles (n) = PV/ RT = 0.009 moles

b) We know that,

Cp / Cv = γ  and

Cp - Cv = R

So, Cv = R / γ-1 = 8.3/0.5 = 16.65

c) Given that,

P1 = 150 kPa = 15 × 10⁴ Pa

P2= ?

V1= 150 cm³ = 150 × 10⁻⁶ m³

γ = 1.5

V2= 50 cm³ = 50 × 10⁻⁶ m³

T1= 300K

T2= ?

Since the process is adiabatic so-

P1 V1 ^γ = P2 V2 ^ γ

From here P2 is found to be 780 k Pa

Again we know that ,

P1 ^ 1-γ T1 ^γ = P2 ^1-γ T2 ^γ

By putting the values we get-

T2 = 520 K

d) From the first law of thermodynamics-

dQ = W + dU or

W= -dU (As the process is adiabatic, so dQ =0)

W= -nCv dT

W= -0.009 × 16.6 × (520 - 300)

W= -33 J

e) We know that,

dU = nCv dT

dU= 0.009 × 16.6 × 220

dU= 33 J

Answer 2

Change in Internal Energy of Gas is 33 J

Explanation:

We know that,

PV = nRT

• And Given P = 150 KPa = $150 \times 10^3 Pa$

• V = $150 cm^3 = 150 \times 10^-^6 m^3$

• T = 300 k

a.) n = $\frac {PV}{RT}$

= $\frac {150 \times 10^3 \times 150 \times 10^-^6}{8.3 \times 300}$

= $9.036 \times 10^-^3$

= 0.009 moles.

b.) $\frac {C_p}{C_v} = \gamma$

$\frac {\gamma R}{(\gamma - 1)C_v} = \gamma\\ (Because C_p = \frac {\gamma R}{\gamma - 1}$

$C_v = \frac {R}{gamma - 1} = \frac {8.3}{1.5 - 1}$

= $\frac {8.3}{0.5} = 2R = 16.6\ Jmole^-^1$

c.) Given $P_1 = 150\ KPa = 150\times10^3\ Pa,\ P_2 = ?$

$V_1 = 150\ cm^3 = 150 \times 10^-^6\ m^3, \gamma = 1.5$

$V_2 = 50\ cm^3 = 50 \times 10^-^6\ m^3, T_1 = 300\ k, T_2$ =?

Since the process is adiabatic, Hence -

$P_1V_{1}^{\gamma} = P_2V_{2}^{\gamma}$

So, $150 \times 10^3(150 \times 10^-^6)^{\gamma} = P_2 \times (50 \times 10^-^6)^{\gamma}$

$P_2 = 150 \times 10^3 \times \frac {150 \times 10^-^6}{50 \times 10^-^6} = 150000 \times 3^{1.5}$

= $779.422 \times 10^3 Pa$ ≈ 780 KPa  

(d)$\Delta Q = W + \Delta U or W = -\Delta U [\Delta$ U = 0, in adiabatic]  

= $-nC_VdT = -0.009 \times 16.6 \times (520-300)$

= $-0.009 \times 16.6 \times 220 = -32.8 J$  ≈ –33 J  

(e) $\Delta U = nC_VdT$

= $0.009 \times 16.6 \times 220$ ≈ 33 J