Question

An ideal monoatomic gas at temperature 27∘C and pressure 106N/m2 cal of heat is added to the system without changing the volume. Calculate the final temperature of the gas. (R=8.31J/mol−KandJ=4.18J/cal)

Answer 1

For n mole of gas, we have PV=nRT

Here, P=106N/m2,V=10L=10−2m3andT=27∘C=300K

So, n=PVRT+=106×10−2(8.31×300)=4

For ‘monoatomic’ gas, CV=3R2

Thus, CV=3×8.31(J/mol−K)2=3×8.31(2×4.18)∼3cal/(mole−K)

Let ΔT be the rise in temperature when n mole of gas is given Q cal of heat at constant volume.

Then, Q=nCVΔT

⇒ΔT=QnCV=10,000cal/[4mole×3cal(mole−K)]=833K

Thus, final temperature of gas is, T+ΔT=300+833K=1133K=860∘C

PLEASE MARK THIS AS THE BRAINLIEST ANSWER ❤️ ❤️