Question

g Suppose a sample of an ideal gas in a container is subjected to a temperature change. A decrease in temperature will the kinetic energy and average speed of the gas particles. As a result, the pressure on the walls of the container will If the gas starts at 25 ∘ C, what temperature would the gas need to reach for its pressure to double? temperature = ∘ C

Answer 1

Answer:

323.15 °C

Explanation:

Considering the ideal gas equation as:

$PV=nRT$

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of  the gas.

P ∝ T

Also,

Using Charle's law  

$\frac {P_1}{T_1}=\frac {P_2}{T_2}$

Given ,  

P₂ = 2P₁

T₁ = 25 °C

T₂ = ?

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

Using above equation as:

$\frac{P_1}{298.15}=\frac{2P_1}{T_2}$

$T_2=2\times 298.15\ K$

New temperature = 596.3 K

Also,

T(K) - 273.15 = T( °C)

So, Temperature = 596.3 - 273.15 °C = 323.15 °C