Question

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.

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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