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ArbitrLikvidat [17]
2 years ago
9

A research balloon at ground level contains 12 L of helium (He) at a pressure of 725mmHg and a temperature of 30.00∘C. When the

balloon has risen to its highest altitude, the volume increases to 28 L and the pressure decreases to 252 mmHg. What will be the temperature of the gas under these conditions?
Chemistry
1 answer:
Helga [31]2 years ago
5 0

<u>Answer:</u> The temperature when the volume and pressure has changed is -27.26°C

<u>Explanation:</u>

To calculate the temperature when pressure and volume has changed, we use the equation given by combined gas law. The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=725mmHg\\V_1=12L\\T_1=30.00^oC=[30+273]K=303K\\P_2=252mmHg\\V_2=28L\\T_2=?K

Putting values in above equation, we get:

\frac{725mmHg\times 12L}{303K}=\frac{252mmHg\times 28L}{T_2}\\\\T_2=\frac{252\times 28\times 303}{725\times 12}=245.74K

Converting this into degree Celsius, we get:

T(K)=T(^oC)+273

245.74=T(^oC)+273\\T(^oC)=-27.26^oC

Hence, the temperature when the volume and pressure has changed is -27.26°C

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Answer:

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Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

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It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

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