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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
- c)(T – 2.4) K
- d)(T + 4) K
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
One mole of an ideal gas at an initial temperature of T K does 6R joul...
We know that for an adiabatic process, $PV^\gamma$ is constant, where $\gamma$ is the ratio of specific heats. Let the initial pressure and volume be $P_i$ and $V_i$, and let the final pressure and volume be $P_f$ and $V_f$. Then we have:
$$P_i V_i^\gamma = P_f V_f^\gamma$$
We also know that the work done by the gas is given by:
$$W = \frac{\gamma}{\gamma - 1} P_i V_i \left(\left(\frac{V_f}{V_i}\right)^{\gamma - 1} - 1\right)$$
Since we are given that the gas does 6R joules of work, we can set $W = 6R$ and solve for $V_f/V_i$:
$$6R = \frac{\gamma}{\gamma - 1} P_i V_i \left(\left(\frac{V_f}{V_i}\right)^{\gamma - 1} - 1\right)$$
$$\frac{V_f}{V_i} = \left(1 + \frac{6R(\gamma - 1)}{\gamma P_i V_i}\right)^{1/(\gamma - 1)}$$
Finally, we can use the ideal gas law to relate the initial and final temperatures:
$$\frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f}$$
Substituting in our expressions for $P_i V_i$, $P_f V_f$, and $V_f/V_i$, we get:
$$\frac{T_f}{T_i} = \frac{P_i V_i}{P_f V_f} = \left(\frac{V_i}{V_f}\right)^{\gamma - 1} = \left(1 + \frac{6R(\gamma - 1)}{\gamma P_i V_i}\right)^{-(\gamma - 1)/\gamma}$$
Substituting in $\gamma = 5/3$ and simplifying, we get:
$$\frac{T_f}{T_i} = \left(1 - \frac{2}{5}\frac{R}{C_V T_i}\right)$$
where $C_V$ is the specific heat at constant volume. Since the gas is ideal and monoatomic, we have $C_V = \frac{3}{2}R$, so we can simplify further:
$$\frac{T_f}{T_i} = \left(1 - \frac{4}{15}\frac{1}{T_i}\right)$$
Thus, the final temperature is:
$$T_f = T_i\left(1 - \frac{4}{15}\frac{1}{T_i}\right) = \frac{11}{15}T_i$$
Therefore, the answer is (d) $\frac{11}{15}T$.
$$P_i V_i^\gamma = P_f V_f^\gamma$$
We also know that the work done by the gas is given by:
$$W = \frac{\gamma}{\gamma - 1} P_i V_i \left(\left(\frac{V_f}{V_i}\right)^{\gamma - 1} - 1\right)$$
Since we are given that the gas does 6R joules of work, we can set $W = 6R$ and solve for $V_f/V_i$:
$$6R = \frac{\gamma}{\gamma - 1} P_i V_i \left(\left(\frac{V_f}{V_i}\right)^{\gamma - 1} - 1\right)$$
$$\frac{V_f}{V_i} = \left(1 + \frac{6R(\gamma - 1)}{\gamma P_i V_i}\right)^{1/(\gamma - 1)}$$
Finally, we can use the ideal gas law to relate the initial and final temperatures:
$$\frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f}$$
Substituting in our expressions for $P_i V_i$, $P_f V_f$, and $V_f/V_i$, we get:
$$\frac{T_f}{T_i} = \frac{P_i V_i}{P_f V_f} = \left(\frac{V_i}{V_f}\right)^{\gamma - 1} = \left(1 + \frac{6R(\gamma - 1)}{\gamma P_i V_i}\right)^{-(\gamma - 1)/\gamma}$$
Substituting in $\gamma = 5/3$ and simplifying, we get:
$$\frac{T_f}{T_i} = \left(1 - \frac{2}{5}\frac{R}{C_V T_i}\right)$$
where $C_V$ is the specific heat at constant volume. Since the gas is ideal and monoatomic, we have $C_V = \frac{3}{2}R$, so we can simplify further:
$$\frac{T_f}{T_i} = \left(1 - \frac{4}{15}\frac{1}{T_i}\right)$$
Thus, the final temperature is:
$$T_f = T_i\left(1 - \frac{4}{15}\frac{1}{T_i}\right) = \frac{11}{15}T_i$$
Therefore, the answer is (d) $\frac{11}{15}T$.
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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) Kb)(T + 2.4) Kc)(T – 2.4) Kd)(T + 4) KCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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) Kb)(T + 2.4) Kc)(T – 2.4) Kd)(T + 4) KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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) Kb)(T + 2.4) Kc)(T – 2.4) Kd)(T + 4) KCorrect answer is option 'A'. Can you explain this answer?.
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) Kb)(T + 2.4) Kc)(T – 2.4) Kd)(T + 4) KCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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) Kb)(T + 2.4) Kc)(T – 2.4) Kd)(T + 4) KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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) Kb)(T + 2.4) Kc)(T – 2.4) Kd)(T + 4) KCorrect answer is option 'A'. Can you explain this answer?.
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