Question

A gas at 37 $^\circ$C is compressed adiabatically to half of its volume, then the final temperature of the gas is (Ratio of specific heat capacities of the gas is 1.5)

• A. 165.3 $^\circ$C
• B. 438.3 $^\circ$C
• C. 400 $^\circ$C
• D. 0 $^\circ$C

Hint:

For an adiabatic process: $T V^{\gamma-1} = \text{constant}$.
Always convert temperatures to Kelvin for thermodynamic calculations involving ratios or products of temperatures.
Convert the final temperature back to Celsius if the options are in Celsius.
$(\text{ratio})^{\text{0.5}} = \sqrt{\text{ratio}}$. Use a suitable approximation for $\sqrt{2}$ (e.g., 1.414).

Answer 1

The Correct Option is A

For an adiabatic process, the relationship between temperature ($T$) and volume ($V$) is given by: $T V^{\gamma-1} = \text{constant}$, or $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$. So, $T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}$. Given values: Initial temperature $T_1 = 37 ^\circ\text{C}$. Convert to Kelvin: $T_1 = 37 + 273.15 \approx 310.15 \text{ K}$. (Using 273 for simpler calculation: $T_1 = 37 + 273 = 310 \text{ K}$). The gas is compressed to half of its volume, so $V_2 = \frac{1}{2}V_1$. This means the ratio $\frac{V_1}{V_2} = 2$. The ratio of specific heat capacities $\gamma = 1.5$. So, $\gamma-1 = 1.5 - 1 = 0.5$. Substitute these values into the formula for $T_2$: $T_2 = T_1 (2)^{\gamma-1} = T_1 (2)^{0.5} = T_1 \sqrt{2}$. Using $T_1 = 310 \text{ K}$ (from $37+273$ for simpler match with options which are likely based on this): $T_2 = 310 \text{ K} \times \sqrt{2}$. Using $\sqrt{2} \approx 1.41421$: $T_2 \approx 310 \times 1.41421 \approx 438.4051 \text{ K}$. To convert $T_2$ back to Celsius: $T_2(^\circ\text{C}) = T_2(\text{K}) - 273 = 438.4051 - 273 = 165.4051 ^\circ\text{C}$. This value is closest to option (a) 165.3 $^\circ$C. If we use $\sqrt{2} \approx 1.414$ (a common 3-decimal approximation): $T_2 \approx 310 \times 1.414 = 438.34 \text{ K}$. $T_2(^\circ\text{C}) \approx 438.34 - 273 = 165.34 ^\circ\text{C}$. This matches option (a) 165.3 $^\circ$C very well when rounded to one decimal place. \[ \boxed{165.3 ^\circ\text{C}} \]