Question

A sample of an ideal gas $\frac{\gamma }{1.5}$ is compressed adiabatically from a volume of 150 $c{m}^3$ to $50c{m}^3$. The initial pressure and the initial temperature are 150 K Pa and 300 K. Find

(a) The number of moles of the gas in the sample

(b) The molar heat capacity at caonstant volume

(c) The final pressure and tempereture

(d) The work done by the gas in the process and

(e) The changr in internal energy of the gas

Answer 1

PV = nRT

Given, P + 150 KPa = 150 $\times {10}^3$ Pa,

V = $150c{m}^3=150\times {10}^{-6}{m}^3$

T = 300 K

(a) n = $\frac{PV}{RT}=9.036\times {10}^{-3}$

= 0.009 moles.

(b) $\frac{Cp}{Cv}=\gamma ,CP-Cv=R$

So, Cv = $\frac{R}{\gamma -1}=\frac{8.3}{0.5}=16.55/moles.$

(c) Given,

$P_1=150KPa=150\times {10}^{3}Pa,$

$P_2=?V_1=150c{m}^3$

= $150\times {10}^{-6}{m}^3$

$\gamma$ = 1.5

$V_2=50c{m}^3=50\times {10}^{-6}{m}^3$

$T_1=300K,T_2=?$

Since the process is adiabatic hence-

$P_1{V}_1\gamma =P_2{V}_2\gamma$

$\Rightarrow 150\times {10}^3\times (150\times {10}^{-6})\gamma$

= $p_2\times (50\times {10}^{-6})\gamma$

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

= $150000\times (3{)}^{1.5}$

= $779.422\times {10}^{3}Pa$

= 780 KPa

Again,

$P_1^{1-\gamma }{T}_1^{\gamma }=P_1^{1-\gamma }{T}^2\gamma$

$\Rightarrow {(150\times {10}^3)}^{1-1.5}\times {\left(330\right)}^{1.5}$

= ${(780\times {10}^3)}^{1-1.5}\times T_2^{1.5}$

$\Rightarrow T_2^{1.5}={(150\times {10}^3)}^{1-1.5}\times (300{)}^{1.5}\times ({300}^{1.5})$

= 11849.050

$\Rightarrow T_2=(11849.050{)}^{\frac{1}{5}}$

= 519.74 = 520

(d) dQ = W + dU

or W = -dU [dQ = 0, in adiabatic]

= -nCvdT

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

= $-0.009\times 16.6\times 220$

= -32.8 J = -33 J

(e) dU = nCvdT

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