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

Step-by-step explanation:

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

`(P_1V_1)/(T_1)=(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:

`(725mmHg* 12L)/(303K)=(252mmHg* 28L)/(T_2)\\\\T_2=(252* 28* 303)/(725* 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