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

178 torr

Step-by-step explanation:

Using Ideal gas equation for same mole of gas as

`\frac {{P_1}* {V_1}}{T_1}=\frac {{P_2}* {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}* {V_1}}{T_1}=\frac {{P_2}* {V_2}}{T_2}`

`\frac {{95.0}* {5}}{301.15}=\frac {{P_2}* {5}}{563.15}`

Solving for V₂ , we get:

P₂ = 178 torr