Question

A sample of an ideal gas (5.00 L) in a closed container at 28.0°C and 95.0 torr is heated to 290 °C. The pressure of the gas at this temperature is torr. О 178 ООО О 50.8

Answer 1

Answer:

178 torr

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,  

V₁ = 5 mL

V₂ = 5 mL (Closed container)

P₁ = 95.0 torr

P₂ = ?

T₁ = 28.0 ºC

T₂ = 290 ºC

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (28 + 273.15) K = 301.15 K  

T₂ = (290 + 273.15) K = 563.15 K  

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{95.0}\times {5}}{301.15}=\frac {{P_2}\times {5}}{563.15}$

Solving for V₂ , we get:

P₂ = 178 torr