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Step-by-Step Solution
Step 1: Recognize the Process
This is a problem involving an adiabatic compression of a diatomic ideal gas (air). In an adiabatic process (where no heat is exchanged with the surroundings), the relationship between temperature and volume for an ideal gas is given by:
$T_1\,V_1^{\gamma - 1} \;=\; T_2\,V_2^{\gamma - 1}$
Here, $T_1$ and $V_1$ are the initial temperature and volume, while $T_2$ and $V_2$ are the final temperature and volume. The exponent $\gamma$ is the ratio of specific heats, $C_p/C_v$.
Step 2: Collect the Known Information
Initial temperature: $T_1 = 300\,\text{K}$
Final volume: $V_2 = \dfrac{1}{32} \,V_1$ (which means the gas is compressed to 1/32 of its original volume)
For a diatomic ideal gas (like air), $\gamma = \dfrac{7}{5} = 1.4$
Step 3: Apply the Adiabatic Relation
Using $T_1\,V_1^{\gamma - 1} = T_2\,V_2^{\gamma - 1}$, we solve for $T_2$:
$$
T_2 \;=\; T_1 \,\left(\dfrac{V_1}{V_2}\right)^{\gamma - 1}.
$$
Step 4: Substitute the Values
Since $V_2 = \dfrac{1}{32}\,V_1$, we have $V_1/V_2 = 32$. Therefore,
$$
T_2 = 300 \times 32^{\,\gamma - 1}.
$$
Given $\gamma - 1 = 1.4 - 1 = 0.4$, and $32 = 2^5$, we get:
$$
32^{\,0.4} = (2^5)^{\,0.4} = 2^{\,5 \times 0.4} = 2^2 = 4.
$$
So,
$$
T_2 = 300 \times 4 \;=\; 1200\,\text{K}.
$$
Step 5: Calculate the Change in Temperature
The change in temperature, $\Delta T$, is given by:
$$
\Delta T = T_2 - T_1 = 1200\,\text{K} - 300\,\text{K} = 900\,\text{K}.
$$
Final Answer
The change in temperature of the gas during the adiabatic compression is $900\,\text{K}$.
