air at 1.02 bar 22 degree certificate initially occupies a cylinder volume of .015m³ is compressed isentropically by the piston to a pressure of 0.8 bar. Determine the final temperature, final volume and work done ?
Answers
Answer:
Explanation:
2.751 j
Initial pressure, p_{1} = 1.02p
1
=1.02 bar
Initial temperature, T_{1} = 22 + 273 = 295T
1
=22+273=295 K
Initial volume, V_{1} = 0.015 m^{3}V
1
=0.015m
3
Final pressure, p_{2} = 6.8p
2
=6.8 bar
Law of compression : pv^{\gamma } = Cpv
γ
=C
(i) Final temperature :
Using the relation,
\frac{T_{2}}{T_{1}}\left ( \frac{P_{2}}{P_{1}} \right )^{\frac{\gamma -1}{\gamma }}
T
1
T
2
(
P
1
P
2
)
γ
γ−1
\frac{T_{2}}{295}=\left ( \frac{6.8}{1.02} \right )^{\frac{1.4-1}{1.4}}
295
T
2
=(
1.02
6.8
)
1.4
1.4−1
[\because γ for \, air = 1.4][∵γforair=1.4]
\therefore∴ T_{2}295\left ( \frac{6.8}{1.02} \right )\frac{1.4-1}{1.4}= 507.24T
2
295(
1.02
6.8
)
1.4
1.4−1
=507.24 K
i.e. Final temperature = 507.24 – 273 = 234.24°C507.24–273=234.24°C .
(ii) Final volume :
Using the relation,
p_{1}V_{1}^{\gamma }=P_{2}V_{2}^{\gamma }p
1
V
1
γ
=P
2
V
2
γ
\frac{P_{1}}{P_{2}}=\left ( \frac{V_{2}}{V_{1}} \right )^{\gamma }
P
2
P
1
=(
V
1
V
2
)
γ
or \frac{V_{2}}{V_{1}}=\left ( \frac{P_{1}}{P_{2}} \right )^{\frac{1}{\gamma }}
V
1
V
2
=(
P
2
P
1
)
γ
1
\therefore∴ V_{2}=V_{1}\times \left ( \frac{P_{1}}{P_{2}} \right )^{\frac{1}{\gamma }}=0.015\times \left ( \frac{1.02}{6.8} \right )^{\frac{1}{1.4}}=0.00387 m^{3} V
2
=V
1
×(
P
2
P
1
)
γ
1
=0.015×(
6.8
1.02
)
1.4
1
=0.00387m
3
i.e., Final volume = 0.00387 m^{3} 0.00387m
3
.
(iii) Wond done :
Now, work done on the air,
W=\frac{mR\left ( T_{1}-T_{2} \right )}{\left ( \gamma -1 \right )}W=
(γ−1)
mR(T
1
−T
2
)
…(i)
where m is the mass of air and is found by the following relation
pV = mRT
\therefore∴ m=\frac{p_{1}V_{1}}{RT_{1}}=\frac{1.02\times 10^{5}\times 0.015}{0.287\times 10^{3}\times 295}m=
RT
1
p
1
V
1
=
0.287×10
3
×295
1.02×10
5
×0.015
\left [ \because R for \, air = 0.287 ×10^{3} \right ][∵Rforair=0.287×10
3
]
= 0.01807 kg
\therefore∴ W=\frac{0.01807\times 0.287\times 10^{3}\left ( 295-507.24 \right )}{\left ( 1.4-1 \right )}=-2751J or – 2.751W=
(1.4−1)
0.01807×0.287×10
3
(295−507.24)
=−2751Jor–2.751 kJ
i.e., Work done = 2.751 kJ.
(–ve sign indicates that work is done on the air).