Question

A 9.80-L container holds a mixture of two gases at 53

Answer #1

**given**

**volume of container = 9.8**

**temperature = 53 C = 53 + 273 = 326 K**

**now**

**total pressure = partial pressure of A (PA) + partial
pressure of B (PB)**

**so**

**total pressure = 0.165 + 0.829**

**total pressure (P) = 0.994 atm**

**we know that**

**PV = nRT**

**so**

**0.994 x 9.80 = n x 0.0821 x 326**

**n = 0.364**

**so**

**the total moles = 0.364**

**now**

**0.18 moles of third gas is added**

**so**

**new total moles = 0.364 + 0.18 = 0.544**

**also given**

**volume and temperature remains constant**

**so use**

**PV = nRT**

**P x 9.8 = 0.544 x 0.0821 x 326**

**P = 1.4856**

**so the total pressure becomes 1.4856 atm**

A 20.0 L container at 303 K holds a mixture of two gases with a
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atm

At a certain temperature, a 14.0-L container holds four gases in
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