Question

A sample of 500 mL Ne gas is at 0.868 atm and 55 C. What is the new volume if the pressure is increased to 1.35 atm and the temperature is decreased to 25 C

Answer 1

Answer:

292.10 mL

Explanation:

From the question we are given;

• Initial volume, V1 of Ne gas is 500 mL
• Initial pressure, P1 of Ne gas is 0.868 atm
• Initial temperature, T1 of Ne gas is 55°C
• but, K = °C + 273.15, thus, T1 = 328.15 K
• Final pressure of the gas, P2 = 1.35 atm
• Final temperature of the gas, T2 = 25°C

                                                              = 298.15 K

We are required to calculate the new volume, V2 of the gas;

Using the combined gas equation;

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Rearranging the formula we can calculate the new volume, V2;

$V2=\frac{P1V1T2}{P2T1}$

$V2=\frac{(0.868atm)(500mL)(298.15K|)}{(1.35atm)(328.15K)}$

$V2=292.10mL$

Therefore, the new volume is 292.10 mL