Question

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?

Answer 1

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

Explanation:

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