Question

For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temperature of a sample of ammonia gas is raised from −25.0°C⁢ to 16.0°C, and at the same time the pressure is changed. If the initial pressure was 0.29kPa and the volume increased by 50.0%, what is the final pressure? Round your answer to the correct number of significant digits.

Answer 1

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

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

where,

$P_1\text{ and }V_1$ are initial pressure and volume at initial temperature $T_1$.

$P_2\text{ and }V_2$ are final pressure and volume at initial temperature $T_2$.

We are given:

$P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V$

$T_1=-25^oC=248.15 K$

$T _2 = 16^oC=289.15 K$

Putting values in above equation, we get:

$\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}$

$P_1=0.225 kPa$

Hence, the new pressure will be 0.225  kPa.