SATHEE: Thermodynamics - Result Question 27

Thermodynamics - Result Question 27

28. One mole of an ideal gas at an initial temperature of $T K$ does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$ , the final temperature of gas will be

[2004]

(a) $(T - 4) K$

(b) $(T + 2.4) K$

(d) $(T + 4) K$

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Answer:

Correct Answer: 28. (a)

Solution:

(a) $T_{1} = T, W = 6 R$ joules, $γ = \frac{5}{3}$

$\begin{aligned} W & = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = \frac{n R T_{1} - n R T_{2}}{γ - 1} \\ = \frac{n R (T_{1} - T_{2})}{γ - 1} \end{aligned}$

$n = 1, T_{1} = T \Rightarrow \frac{R (T - T_{2})}{5 / 3 - 1} = 6 R$

$\Rightarrow T_{2} = (T - 4) K$

Thermodynamics - Result Question 27

28. One mole of an ideal gas at an initial temperature of $T K$ does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$ , the final temperature of gas will be

Answer:

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Thermodynamics - Result Question 27

28. One mole of an ideal gas at an initial temperature of TK does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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28. One mole of an ideal gas at an initial temperature of $T K$ does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$ , the final temperature of gas will be