The pressure of 250 kPa acting on 15 cubic meters of gas at a temperature of 100 K is increased to 500 kPa and the volume is increased to 30 cubic meters. What is the temperature of the gas at this new pressure and volume?

Question

Luna Chen · Accepted Answer

The final temperature will be 400 K.

This is an example of the combined gas law. The equation to use is #(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#. Given Initial pressure, #P_1="250 kPa"# Initial volume, #V_1="15 m"^3"# Initial temperature, #T_1="100 K"# Final pressure, #P_2="500 kPa"# Final volume, #V_2="30 m"^3"# Unknown Final temperature, #T_2# Solution Rearrange the equation to isolate #T_2#. Substitute the given values into the equation and solve. #(P_1V_1)/(T_1)=(P_2V_2)/(T_2)# #T_2=(T_1P_2V_2)/(P_1V_1)# #T_2=(100"K"·500cancel"kPa"·30cancel"m"^3)/(250cancel"kPa"·15cancel"m"^3)="400K"#

Mateo Aguilar · Answer

undefined Use the ideal gas law: $PV = nRT$. Rearrange it to find the final temperature ($T_2$):

$$T_2 = \frac{P_2 \cdot V_2 \cdot T_1}{P_1 \cdot V_1}$$

Substitute values:

$$T_2 = \frac{(500 \, \text{kPa}) \cdot (30 \, \text{m}^3) \cdot (100 \, \text{K})}{(250 \, \text{kPa}) \cdot (15 \, \text{m}^3)}$$

Calculate $T_2$.