Question

One mole of an ideal gas is at temperature \( T \) K. The \( \gamma \) value of this gas is \( \frac{5}{3} \). Now the gas does 12R Joules of work adiabatically (R is the universal gas constant). Then the final temperature of the gas will be:

• A. \( T - 8 \, \text{K} \)
• B. \( T + 4 \, \text{K} \)
• C. \( T - 4.4 \, \text{K} \)
• D. \( T - 6 \, \text{K} \)

Hint:

In adiabatic processes, the relationship between temperature and work done is crucial to finding the final temperature.

Answer 1

The Correct Option is A

To solve this problem, we need to use the formula for the change in temperature during an adiabatic process. The formula for the work done \( W \) during an adiabatic process is:

\[ W = nC_v(T_i - T_f) \]

where \( n \) is the number of moles, \( C_v \) is the molar heat capacity at constant volume, \( T_i \) is the initial temperature, and \( T_f \) is the final temperature.

We know \( \gamma = \frac{C_p}{C_v} \), and for one mole of an ideal gas, \( C_p - C_v = R \). Given that \( \gamma = \frac{5}{3} \), we can calculate \( C_v \) as follows:

\[ \gamma = \frac{C_p}{C_v} = \frac{C_v + R}{C_v} = \frac{5}{3} \]

Solving for \( C_v \):

\[ \frac{C_v + R}{C_v} = \frac{5}{3} \]

\[ 3(C_v + R) = 5C_v \]

\[ 3C_v + 3R = 5C_v \]

\[ 2C_v = 3R \]

\[ C_v = \frac{3}{2}R \]

Plugging this value back into the formula for work done:

\[ 12R = 1 \times \frac{3}{2}R(T_i - T_f) \]

\[ 12 = \frac{3}{2}(T - T_f) \]

\[ 24 = 3(T - T_f) \]

\[ T - T_f = 8 \]

Thus, the final temperature of the gas is:

\[ T_f = T - 8 \, \text{K} \]

This matches the provided answer \( T - 8 \, \text{K} \).

Answer 2

For an adiabatic process, the relation between temperature and work done is given by: \[ W = \frac{P_{\text{final}} V_{\text{final}} - P_{\text{initial}} V_{\text{initial}}}{1 - \gamma} \] For a monoatomic ideal gas with \( \gamma = \frac{5}{3} \), the final temperature \( T_f \) is given by: \[ T_f = T_i - \frac{8R}{m} \] Thus, the final temperature will be \( T - 8 \, \text{K} \).

Answer 3

This question asks for the final temperature of one mole of an ideal gas with \( \gamma = \frac{5}{3} \) after it does \( 12R \) joules of work adiabatically, starting from temperature \( T \). Instead of deriving \( C_v \) from the ratio equations directly, we use the direct formula \( C_v = \frac{R}{\gamma - 1} \), which follows from combining \( C_p - C_v = R \) with \( \gamma = \frac{C_p}{C_v} \).

With \( \gamma = \frac{5}{3} \):

\[ C_v = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2}R \]

For an adiabatic process, no heat is exchanged, so the work done by the gas comes entirely at the expense of its internal energy:

\[ W = -\Delta U = nC_v(T_i - T_f) \]

With \( n = 1 \), \( W = 12R \), and \( C_v = \frac{3}{2}R \):

\[ 12R = \frac{3}{2}R(T - T_f) \]

\[ T - T_f = \frac{12R}{\frac{3}{2}R} = 8 \]

\[ T_f = T - 8 \, \text{K} \]

Now checking each option against this result:

1. Option A, \( T - 8 \, \text{K} \): this is exactly the value obtained above from \( T - T_f = 8 \). This option is consistent with the calculation.
2. Option B, \( T + 4 \, \text{K} \): this would mean the gas's temperature rose while it did positive work adiabatically, which is impossible — a gas doing work on its surroundings with no heat input can only cool down, never heat up. This option is ruled out on physical grounds alone.
3. Option C, \( T - 4.4 \, \text{K} \): for this to be true, \( C_v \) would have to equal \( \frac{12R}{4.4} \approx 2.73R \), which does not correspond to \( \gamma = \frac{5}{3} \) for any consistent value of \( C_v \). This option is inconsistent with the given \( \gamma \).
4. Option D, \( T - 6 \, \text{K} \): this would require \( C_v = \frac{12R}{6} = 2R \), giving \( \gamma = \frac{C_v+R}{C_v} = \frac{3R}{2R} = 1.5 \), not \( \frac{5}{3} \). This option also fails to match the given \( \gamma \).

Only option A survives the check against both the energy balance and the given value of \( \gamma \).

Therefore, the correct answer is \( T - 8 \, \text{K} \).