Question #134380
An ideal gas at 10.3 °C and a pressure of 2.48 x 10^5 Pa occupies a volume of 3.12 m3. (a) How many moles of gas are present? (b) If the volume is raised to 5.45 m3 and the temperature raised to 25.6 °C, what will be the pressure of the gas?
1
Expert's answer
2020-09-25T09:09:47-0400

The ideal gas equation is given by

pV=nRTpV=nRT ….........(1)

where pp is the pressure in PaPa

VV is the volume in m3m^3

n is the number of mole

R=8.314 J.mol1.K1R = 8.314\ J.mol^{-1}.K^{-1} is the universal gas constant and

TT is the absolute temperature in KK of the gas.


(a) Given, p=2.48×105 Pa, V=3.12 m3,T=10.3°C=(273+10.3)K=283.3Kp=2.48\times 10^5\ Pa,\ V=3.12\ m^3, T=10.3 \degree C=(273+10.3)K=283.3K

From Eq.(1), n=pVRTn=\frac{pV}{RT}

Substituting the given values of p,V,R,Tp,V,R,T, we get

n=2.48×105×3.128.314×283.3 moln=\frac{2.48\times 10^5\times 3.12}{8.314\times 283.3}\ mol

n=328.5 mol\therefore n=328.5\ mol


(b) New volume V=5.45 m3V'=5.45\ m^3

New temperature T=25.6°C=(273+25.6)K=298.6KT'=25.6\degree C=(273+25.6)K=298.6K

The new pressure is given by

p=nRTVp'=\frac{nRT'}{V'}

p=328.5×8.314×298.65.45 Pap'=\frac{328.5\times 8.314\times 298.6}{5.45}\ Pa

p=1.50×105 Pa\therefore p'=1.50\times 10^5\ Pa


Answer: (a) There are 328.5 moles of gas.

(b) The new pressure of the gas will be 1.50×105 Pa1.50\times 10^5\ Pa .


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