Question 668: Thermodynamics & KTG

Question

Temperature remaining constant, the pressure of gas is decreased by $20 \%$. The percentage change in volume

Accepted Answer

When the temperature is held constant and the pressure of a gas is decreased by 20%, we want to find the percentage change in volume. To solve this, we use the ideal gas law relationship, $ P_1 V_1 = P_2 V_2 $.
Given:
The initial pressure ($ P $) is reduced by 20%, resulting in the new pressure ($ P' $) being 80% of the original pressure. Therefore, $ P' = \frac{80}{100} P $.
By substituting into the equation and solving for the volume change:
$  PV = \left(\frac{80}{100} P ight) \cdot V'  $
$  V' = \frac{100}{80} V  $
The percentage increase in volume is calculated by:
$  	ext{Percentage increase in volume} = \left(\frac{V' - V}{V}ight) 	imes 100  $
$  = \left(\frac{\frac{100}{80} V - V}{V}ight) 	imes 100  $
$  = \left(\frac{\frac{100V - 80V}{80}}{V}ight) 	imes 100  $
$  = \left(\frac{20}{80}ight) 	imes 100  $
$  = \frac{1}{4} 	imes 100 = 25\%  $
Thus, the volume increases by 25%.

Answer

increases by $29 \%$

Answer

decreases by $20 \%$

Answer

increases by $25 \%$

Answer

decreases by $25 \%$

Thermodynamics & KTG

Temperature remaining constant, the pressure of gas is decreased by 20 \%. The percentage change in volume

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