Question

An ideal gas at pressure 2.5 × 105 Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure (b) the final temperature and (c) the work done by the gas in the process. Take γ = 1.5

Answer 1

Answer:

Explanation:

P1 = 2.5 × 10P1=2.5×105Pa,

T1=300K

V1=100C,

(a) P1Vγ1=P2Vγ2

implies 2.5×105×V1.5=(V2)1.5×P2

implies P2=7.07×105

=7.1×105

(b) T1Vγ−11=T2Vγ−12

implies300×(100)1.5−1=T2×(1002)1.5−1

=T2×(50)1.5−1

implies 300×10=T2×7.07

T2=424.32K=424K.

(c) Work done by the gas in the process

W=mRγ−1[T2−T1]

=P1V1γ−1[T2−T1]

=2.5×10300×0.5×124

=20.67=21J5 Pa, V1 = 100 cc, T1 = 300 k

Answer 2

(a) The final pressure is $7.1 \times 10^{5} \mathrm{Pa}$

(b) The final temperature is $424 K$

(c) The work done by the gas in the process is $-20.67 \approx-21 J$

Explanation:

Initial Gas pressure, $P_{1}=2.5 \times 10^{5} P a$

Initial gas temperature, $T_{1}=300 K$

Initial gas volume, $V_{1}=100 c c$

(a) That cycle is adiabatic,

$P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$

$2.5 \times 10^{5} \times V^{1.5}=\left(\frac{V}{2}\right)^{1.5} \times P_{2}$

$P_{2}=7.07 \times 10^{5}$

$=7.1 \times 10^{5} \mathrm{Pa}$

(b) To an adiabatic method, too,

$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$

$300 \times(100)^{1.5-1}=T_{2} \times\left(\frac{100}{2}\right)^{1.5-1}$

$=T_{2} \times(50)^{1.5-1}$

$300 \times 10=T_{2} \times 7.07$

$T_{2}=424.32 K=424 K$

(c) Work done through the gas process,

$w=\frac{n R}{\gamma-1}\left[T_{1}-T_{2}\right]$

$=\frac{P_{1} V_{1}}{T(\gamma-1)}\left[T_{1}-T_{2}\right]$

$=\frac{2.5 \times 10}{300 \times 0.5} \times(-124)$

$=-20.67 \approx-21 J$