Answer to Question #138534 in General Chemistry for mena

Question

A sample of gas at 25.00oC has a pressure of 92.50kPa. What temperature would change its pressure to 110.0kPa?

Answer 1

We will use the relationship;

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

From the question it seems Volume is kept constant. When that is the case, the relationship becomes;

$\dfrac{P1}{T1}=\dfrac{P2}{T2}$ (at constant volume)

P1=92.50kPa

P2 = 110.0kPa

T1 = 25 + 273 = 298K

T2 = ?

Solving for T2, the expression changes to;

$T2 = \dfrac{P2 x T1}{P1}$

$T2 = \dfrac{110.0kPa x 298K}{95.50kPa} = 343.246K$

Changing to ^oC, T2 = 70.25 ^oC