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The volume of a gas is increased by 20% from its initial value of \small 80cm^3 without changing pressure. To what temperature the gas needs to be heated if its initial temperature (in oC) is \small 25^{\circ}C?

Option: 1

80.7


Option: 2

78.3


Option: 3

84.6


Option: 4

87.3


Answers (1)

best_answer

20% increase in volume \frac{80\times20}{100}=16cm^3

Thus, the final volume of the gas is 80+16=96cm^3Now, we have:
V_1=80cm^3,V_2=96cm^3

T_1=298K

Thus, according to Charles's law, we have:

\mathrm{\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}}}

\\\mathrm{\therefore{T_{2}} =\frac{V_{2}\, \times \, T_{1}}{T_{2}}} \\ \\ \mathrm{T_{2}=\: \frac{96\, \times \, 298}{80}\: =\: 357.6\ K\: or\: 84.6^{0}C}

 

Posted by

rishi.raj

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