Question

1 litre of an ideal gas (γ = 1.5) at 300 K is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is 100 kPa, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.

Answer 1

Ratio of the final pressure to the initial pressure = 2√2 , work done by the gas =−82 J, change in internal energy = 82 J, final temperature  = 424 K,  work done during the process = −41.4 J, work done by the gas = 103 and total work done in the cycle = −20.4 J

Explanation:

Given data:

γ=1.5

Temperature =300 K

Initial volume "Vi" = 1L,

Final volume "V2" = 1/2L

• (a) The process is adiabatic because volume is suddenly change

P1V1^γ=P2V2^γ

P2 = P1(V1/V2)^γ = P1(2)^γ

OR

P2/P1 = (2)^1.5 = 2√2

• (b) P1 = 100  KPa= 10^5 Pa  

and P2=2√2×10^5 Pa

Work done by adiabatic process

=P1V1 − P2V2/γ−1

=10^5×10^−3−2√2 ×10^5 × 1/2×10^−3/15−1

=−82 J

• (c) Internal energy

dQ=0

dU = −dW = −(−82 J) = 82 J

• (d) T1V1^γ−1 = T2V2^γ−1

T2=T1(V1/V2)^γ−1

= 300(2)^0.5

=300 × √2 = 300 × 1.414

=T2 = 424 K

• (e) The pressure is kept constant. The process is isobaric work done =nRdT

Here: n=PV/RT=10^5×10^−3/R×300 = 1/3 R

So work done =1/3 R × R(300−424)

=− 41.4 J.

• (f) V1/T1 = V2/T2

V1 = V2T/1T2

Work done in this process

= nRTln (V1/V2)

= 1/3 R×R×T×In2

= 100×ln2

= 100×1.039

= 103

• (g) Net work done

= −82−41.4+103

= −20.4 J

Answer 2

(a) 2$\sqrt{2}$

(b) W=82 J

(c) dU = 82 J

(d) $\mathrm{T}_{2}=424 \mathrm{K}$

(e) Work done = - 41.4 J

(f) Work done = 103 J  

(g) Total work done = - 20.4 J  

Explanation:

Given data in the question  

Temperature T = 300 K

Initial gas  volume ,$V_1 = 1 L$

Final gas volume,$V_2 = \frac{1}{2} L$

(a) The process is adiabatic due to the sudden change in volume; therefore, no heat exchange is allowed.

$\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}$

$\mathrm{P}_{2}=\mathrm{P}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{V}_{2}}\right)^{\gamma}=\mathrm{P}_{1} 2^{\gamma}$

$\frac{P_{2}}{P_{1}}=2^{1.5}=2 \sqrt{2}$

(b)

$P_1 = 100 kPa = 105 Pa$

$\text { And } P_{2}=2 \sqrt{2} \times 10^{5} \mathrm{Pa}$

Work done through an adiabatic method,

$W=\frac{P_{1} V_{1}-P_{2} V_{2}}{\gamma-1}$

$W=\frac{10^{5} \times 10^{-3}-2 \sqrt{2} \times 10^{5} \times \frac{1}{2} \times 10^{-3}}{1.5-1}$

$W=\frac{10^{2}-\sqrt{2} \times 10^{2}}{0.5}$

$W=\frac{10^{2}-1.41 \times 10^{2}}{0.5}$

$W=\frac{-0.41 \times 10^{2}}{0.5}$

W=82 J

(c)  in case of Internal energy,

dQ = 0, as it is an adiabatic method  

dU = - dW = - (- 82 J)

dU = 82 J

(d) To an adiabatic method,

$\mathrm{T}_{1} \mathrm{V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{V}_{2}^{\gamma-1}$

$\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{V}_{2}}\right)^{\gamma-1}$

$\mathrm{T}_{2}=300 \times 2^{0.5}$

$\mathrm{T}_{2}=300 \times 2 \sqrt{2}=300 \times 1.41$

$\mathrm{T}_{2}=424 \mathrm{K}$

(e) Pressure remains constant.

The mechanism is isobaric; thus, work is done = P∆V=nRdT.

n=PV/RT

$n=\frac{10^{5} \times 10^{-3}}{R \times 300}$

$\mathrm{n}=\frac{1}{3 R}$

$\text { work done }=\frac{1}{3 R} \times R \times(300-424)$

$work done =-41.4 J$

Pressure = constant  

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$V_{1}=V_{2} \frac{V_{1}}{T_{1}}$

(f) Work done in an isothermal method

$W=n R T \ln \frac{V_{2}}{V_{1}}=\frac{1}{3 R} \times R \times T \times {In}(2)$

$W=100 \times \ln 2=100 \times 1.039$

W = 103 J  

(g) total  work done (using thermodynamics first law )

W = - 82 - 41.4 + 103

W = - 20.4 J

Thus the given problem has been calculated