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Marta_Voda [28]
3 years ago
6

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
Chemistry
1 answer:
liq [111]3 years ago
5 0

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

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