Question

# A gas occupies $$500\ cm^{3}$$ at normal temperature. At what temperature will the volume of the gas be reduced by $$20\%$$ of its original volume, pressure being constant?

Solution

## V = $$500 cm^3$$Normal temperature, t = $$0^{\circ}C$$ = 0 + 273 K$$V_{1}$$ = Reduced volume + 20% of $$500 cm^3$$= $$\dfrac{20 \times 500}{100} = 100 cm^3$$Net, $$V_{1}$$ = 500 - 100 = $$400 cm^3$$$$T_{1}$$ = ?P = $$P_{1}$$From gas equation,$$\dfrac{PV}{T} = \dfrac{P_1V_1}{T_1}$$$$\dfrac{P \times 500}{273} = \dfrac{P \times 400}{T_1}$$$$T_1 = \dfrac{273 \times 4}{5}$$        = $$218.4$$        = $$218.4 - 273$$        = $$54.60^{\circ}C$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More