Question

A sample of methane gas, ch4(g), occupies a volume of 60.3 l at a pressure of 469 torr and a temperature of 29.3°c. what would be its temperature at a pressure of 243 torr and volume of 60.3 l?

Answer 1

We can solve this equation using Gay-Lussac's law where it relates the pressure and temperature.
Gay-Lussac's gas law states that pressure is proportional to temperature when the volume is constant.
 $\frac{P}{T} = k$
where P - pressure, T - temperature and k - constant.
for this question we can use this law as volume is constant in both instances.
Therefore 
$\frac{P1}{T1} = \frac{P2}{T2}$
where left side parameters are for the first instance and right side parameters are for the second instance.
$\frac{469 Torr}{29.3} = \frac{243 Torr}{x}$
where x is the temperature we need to calculate
Therefore 
x = 243 * 29.3 / 469  = 15.2 °C
the new temperature is then 15.2 °C

Answer 2

Answer: The temperature when the volume and pressure has changed is -116.4°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:

$P_1=469torr\\V_1=60.3L\\T_1=29.3^oC=[29.3+273]K=302.3K\\P_2=243torr\\V_2=60.3L\\T_2=?K$

Putting values in above equation, we get:

$\frac{469torr\times 60.3L}{302.3K}=\frac{243torr\times 60.3L}{T_2}\\\\T_2=156.63K=-116.37^oC$

Hence, the temperature when the volume and pressure has changed is -116.4°C