SATHEE: Thermodynamics - Result Question 28

Thermodynamics - Result Question 28

29. An ideal gas at $27^{\circ} C$ is compressed adiabatically to $\frac{8}{27}$ of its original volume. The rise in temperature is $(γ = \frac{5}{3})$

[1999]

(a) $475^{\circ} C$

(b) $402^{\circ} C$

(d) $175^{\circ} C$

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Answer:

Correct Answer: 29. (b)

Solution:

(b) $T = 27^{\circ} C = 300 K$

$γ = \frac{5}{3}; V_{2} = \frac{8}{27} V_{1}; \frac{V_{1}}{V_{2}} = \frac{27}{8}$

From adiabatic process we know that

$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$

$\frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{γ - 1} = (\frac{27}{8})^{\frac{5}{3} - 1}$

$\frac{T_{2}}{T_{1}} = \frac{9}{4} \Rightarrow T_{2} = \frac{9}{4} \times T_{1} = \frac{9}{4} \times 300 = 675 K$

$T_{2} = 675 - 273^{\circ} C = 402^{\circ} C$

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Thermodynamics - Result Question 28

29. An ideal gas at 27∘C is compressed adiabatically to 827 of its original volume. The rise in temperature is (γ=53)

Answer:

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29. An ideal gas at $27^{\circ} C$ is compressed adiabatically to $\frac{8}{27}$ of its original volume. The rise in temperature is $(γ = \frac{5}{3})$