Question

5. An ideal gas occupies a volume of 7.0 L at STP. What is its gauge...

5. An ideal gas occupies a volume of 7.0 L at STP. What is its gauge pressure (in kPa) if the volume is halved and the absolute temperature is doubled? Show working equation with answer. Answer must be in kPa.

Homework Answers

Answer #1

Given that Ideal gas occupies 7.0 L at STP, So using ideal gas law:

PV = nRT

At STP, T = 0 C = 273 K, and P = 1 atm = 1.01325*10^5 Pa

V = Volume of gas = 7.0 L = 7.0*10^-3 m^3,

R = Gas constant = 8.314 , So number of moles will be:

n = PV/RT = 1.01325*10^5*7.0*10^-3/(8.314*273)

n = 0.3125 moles

Step 2: Now Volume is halved and temperature is doubled, So now new pressure will be: (while number of moles remains same)

P'*V' = n*R*T'

V' = 7.0/2 = 3.5 L = 3.5*10^-3 m^3

T' = 273*2 = 546 K

So,

P' = n*R*T'/V' = 0.3125*8.314*546/(3.5*10^-3)

P' = Absolute pressure = 405307.5 Pa

Step 3: Now we know that:

Absolute Pressure = Atmospheric pressure + Gauge Pressure

Gauge Pressure = P' - P

Gauge Pressure = 405307.5 - 1.01325*10^5

Gauge Pressure = 303982.5 = 304.0*10^3 Pa

Gauge Pressure = 304.0 kPa

Let me know if you've any query.

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