Question

The pressure inside a hydrogen-filled container was 2.10 atm at 21 ∘C. What would the pressure be if the container was heated to 88 ∘C ?

Answer 1

Answer:

2.58 atm

Explanation:

P1 = 2.1 atm

T1 = 21 °C = 294 K

T2 = 88 °C = 361 K

P2 = ?

By using the gas laws

P / T = constant keeping the volume constant.

P1 / T1 = P2 / T2

2.10 / 294 = P2 / 361

P2 = 2.58 atm

Thus, the pressure becomes 2.58 atm.

Answer 2

Answer:

2.57832 atm

Explanation:

$P_1$ = Initial pressure = 2.1 atm

$P_2$ = Fianl pressure

$T_1$ = Initial Temperature = (21+273.15) K

$T_2$ = Final Temperature = (88+273.15) K

From Gay-Lussacs law we have

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\Rightarrow P_2=\dfrac{P_1\times T_2}{T_1}\\\Rightarrow P_2=\dfrac{2.1\times (88+273.15)}{21+273.15}\\\Rightarrow P_2=2.57832\ atm$

The final pressure would be 2.57832 atm