An ideal monoatomic gas at temperature 27∘C and pressure 106N/m2 occupies 10L volume. 10,000cal of heat is added to the system without changing the volume. Calculate the change in temperature of the gas. Given:R=8.31J/mol-K and J=4.18J/cal
A
500K
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B
833K
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C
300K
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D
0K
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Solution
The correct option is B833K For n moles of gas, we have PV=nRT Here, P=106N/m2,V=10−2m3 and T=27∘C=300K ∴n=PVRT=106×10−28.31×300=4.0
For monoatomic gas, CV=32R Thus, CV=32×8.31J/mol-K =32×8.314.18≈3cal/mol-K Let ΔT be the rise in temperature when n moles of the gas is given Qcal of heat at constant volume. Then, Q=nCVΔT or ΔT=QnCV =10000cal4.0mole×3cal/mol-K ≈833K