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Math  /  Algebra

QuestionA cylinder with a movable piston contains gas at a temperature of 24.3∘C24.3^{\circ} \mathrm{C}, a volume of 2.09 m32.09 \mathrm{~m}^{3}, and an absolute pressure of 13700 Pa .
What will be its final temperature if the gas is compressed to 0.34 m30.34 \mathrm{~m}^{3} and the absolute pressure increases to 63350 Pa ?
Answer in units of ∘C{ }^{\circ} \mathrm{C}.

Studdy Solution

STEP 1

1. The initial temperature of the gas is 24.3∘C24.3^{\circ} \mathrm{C}.
2. The initial volume of the gas is 2.09 m32.09 \mathrm{~m}^{3}.
3. The initial absolute pressure of the gas is 13700 Pa13700 \, \mathrm{Pa}.
4. The final volume of the gas is 0.34 m30.34 \mathrm{~m}^{3}.
5. The final absolute pressure of the gas is 63350 Pa63350 \, \mathrm{Pa}.
6. The process is assumed to be ideal and follows the combined gas law.

STEP 2

1. Convert the initial temperature to Kelvin.
2. Use the combined gas law to find the final temperature in Kelvin.
3. Convert the final temperature back to Celsius.

STEP 3

Convert the initial temperature from Celsius to Kelvin.
The formula to convert Celsius to Kelvin is: T(K)=T(∘C)+273.15 T(K) = T(^{\circ}C) + 273.15
For the initial temperature: T1(K)=24.3+273.15=297.45 K T_1(K) = 24.3 + 273.15 = 297.45 \, \mathrm{K}

STEP 4

Use the combined gas law to find the final temperature in Kelvin.
The combined gas law is: P1V1T1=P2V2T2 \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
Rearrange to solve for T2 T_2 : T2=P2V2T1P1V1 T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}
Substitute the known values: T2=63350 Pa×0.34 m3×297.45 K13700 Pa×2.09 m3 T_2 = \frac{63350 \, \mathrm{Pa} \times 0.34 \, \mathrm{m}^3 \times 297.45 \, \mathrm{K}}{13700 \, \mathrm{Pa} \times 2.09 \, \mathrm{m}^3}
Calculate T2 T_2 : T2=63350×0.34×297.4513700×2.09 T_2 = \frac{63350 \times 0.34 \times 297.45}{13700 \times 2.09}
T2≈224.85 K T_2 \approx 224.85 \, \mathrm{K}

STEP 5

Convert the final temperature from Kelvin back to Celsius.
The formula to convert Kelvin to Celsius is: T(∘C)=T(K)−273.15 T(^{\circ}C) = T(K) - 273.15
For the final temperature: T2(∘C)=224.85−273.15 T_2(^{\circ}C) = 224.85 - 273.15
T2(∘C)≈−48.3 T_2(^{\circ}C) \approx -48.3
The final temperature of the gas is: −48.3∘C \boxed{-48.3^{\circ} \mathrm{C}}

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