Question

0.0625 mol of an ideal gas is contained within a 1.0000 Lvolume container. Its pressure is 142,868 Pa and temperature is X K . What is X?
(HINT: |X| is near an order of magnitude of 102 K ).

Answer 1

Answer: The temperature of the ideal gas is $2.75\times 10^2K$

Explanation:

To calculate the temperature, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = Pressure of the gas = 142,868 Pa = 142.868 kPa    (Conversion factor: 1 kPa = 1000 Pa)  

V = Volume of gas = 1.0000 L

n = number of moles of ideal gas = 0.0625 moles

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the gas = ?

Putting values in above equation, we get:

$142.868kPa\times 1.0000=0.0625mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T\\\\T=275K=2.75\times 10^2K$

Hence, the temperature of the ideal gas is $2.75\times 10^2K$