Thermal Laws

A closed container of volume 0.02m30.02m^3 contains a mixture of neon and argon gases, at a temperature of 27o27^oC and pressure of 1×1051\times 10^5 N/m2N/m^2. The total mass of the mixture is 2828 gm. If the gram molecular weights of neon and argon are 2020 and 4040 respectively, find the masses of the individual gases in container, assuming them to be ideal.(Universal gas constant R=8.314=8.314 J/mol. K)

হানি নাটস

mNe+mAr=28gmm_{Ne}+m_{Ar}=28gm

PV=(nNe+nAr)RTPV=(n_{Ne}+n_{Ar})RT

PVRT=mNe20+mAr40=105×0.028.314×300\dfrac{PV}{RT}=\dfrac{m_{Ne}}{20}+\dfrac{m_{Ar}}{40}=\dfrac{10^5 \times 0.02}{8.314\times 300}

2mNe+mAr=32.07442m_{Ne}+m_{Ar}=32.0744

mNe=4.0744m_{Ne}=4.0744

mAr=23.926m_{Ar}=23.926

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