You know that a gas in a sealed container has a pressure of 111 kPa at 231°C. What will the pressure be if the temperature rises to 475°C?

Question

Owen Ambrose · Accepted Answer

#P_2=165kPa#

Firstly, because this is a sealed container, the volume will always remain constant; additionally, since the temperature is only rising, the number of moles will also remain constant.

From the ideal gas law: #PV=nRT# we can say that at #V=cte# and #n=cte# the pressure #P# is directly proportional to the temperature #T#.

Therefore, #P=kT or P/T=k#, where #k# is a constant #k=(nR)/V#.

Thus, #(P_1)/(T_1)=(P_2)/(T_2)=>P_2=(P_1xxT_2)/(T_1)#

#P_2=(111kPaxx748cancel(K))/(504cancel(K))=165kPa#

Wyatt Adkins · Answer

undefined To find the pressure at the new temperature, we can use the Gay-Lussac's law, which states that the pressure of a gas is directly proportional to its temperature, provided that the volume and the amount of gas remain constant.

$P_1/T_1 = P_2/T_2$

Given:
$P_1 = 111 \, kPa$
$T_1 = 231°C + 273.15 = 504.15 \, K$ (temperature in Kelvin)
$T_2 = 475°C + 273.15 = 748.15 \, K$ (new temperature in Kelvin)

Using the formula:
$P_2 = \frac{P_1 \times T_2}{T_1}$

$P_2 = \frac{111 \times 748.15}{504.15}$

$P_2 ≈ 164.8 \, kPa$

Therefore, the pressure at the new temperature of 475°C will be approximately 164.8 kPa.