Question

A gas has a volume of 400 mL. at 50.0 °C and 120 kPa. What will be its volume at atemperature of 80.0 °C and a pressure of 180 kPa?

Answer 1

291.43mL

Step-by-step explanation

According to the combined gas law;

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where:

• P₁ and P₂ are the, initial and final pressure, respectively

,

• V₁ and V₂ are the, initial and final volume, respectively

,

• T₁ and T₂ are the ,initial and final temperature, respectively (in Kelvin)

Given the following parameters

`\begin{gathered} P_1=120\text{kPa} \\ P_2=180\text{kPa} \\ V_1=400mL \\ V_2=\text{?} \\ T_1=50.0^0=50+273=323K \\ T_2=80.0^0=80+273=353K \end{gathered}`

Substitute the given parameters into the formula to get the final volume V₂

`\begin{gathered} V_2=(P_1V_1T_2)/(P_2T_1) \\ \end{gathered}`

On substituting these values

`\begin{gathered} V_2=\frac{120\cancel{\text{kPa}}*400mL*353\cancel{K}}{180\cancel{\text{kPa}}*323\cancel{K}} \\ V_2=(16,944,000)/(58,140)mL \\ V_2=291.43mL \end{gathered}`

Hence its volume at a temperature of 80.0 °C and a pressure of 180 kPa is 291.43mL