10 grams of argon gas is compressed isothermally and reversibly at a temperature of 27⁰C from 10 K to 5L. Calculate the work done for this process in multiple of 10² calories.

As we know that-

ΔH=ΔU+W.....(1)

Since temperature is constant, the process is isothermal.

∴ΔU=0

W=−nRTlnV1​V2​​=−2.303×nRTlogV1​V2​​[∵the gas is compressed]

Given that:-

mass of Ar=10g

Atomic mass of Ar=40g

∴n=at. massGiven mass​=4010​=0.25 moles

Temperature (T)=27℃=(273+27)K=300K

Initial volume (V₁​)=10L

Final volume (V₂​)=5L

R=2calK−1mol−1

∴W=−2.303×0.25×2×300×log105​

⇒W=−345.45×(log1−log2)

⇒W=−345.45×(−0.3)=103.635cal[∵log2=0.3(Given)&log1=0]

From equation (1), we get

ΔH=0+103.635=103.635cal

Hence the required answer is 103.635cal =1 x 10² cal