Question

If 1.24 moles of H2 produces a pressure of 1.08 atm in a 4.50 L container at -184°C, what will the temperature (in °C) have to be if an additional 0.613 moles of H2 are added to the container and the pressure increases to 2.32 atm? -184°C -162°C -152°C -145 °C - 128 °C • Previous Next

Answer 1

Solution:-

we will solve this question by using the Ideal Gas equation PV= nRT

Here the given Data

P, = 1.08 atm

n = 1.24 mole

T2 = -184 + 273.15 = 89.15 K

As the volume of the container remains the same So V= constant

So here V and R constant, So we can write

⇒ P ∝ nT

⇒

P___ T,na Tana

Given

n = 1.24 mole + 0.613 mole = 1.853 mole

P = 2.32 atm

Using the above formula, we can write

T 11

⇒  

T2 = 2.32 atm x 89.15 KX 1.24 mole 1.853 mole x 1.08 atm

⇒

T2 = 128.15K

⇒

T2 = 128.15K -273.15

⇒

T, = -144.99 °C -145°C