Question
When the pressure exerted on 1.0L of an ideal gas is quadrupled, and the kelvin temperature is tripled,what does the volume become?
Answers
kat
use combined gas laws
DrRebel
Another way to look at this is ... changes in volume are pressure and temperature sensitive...
=> Increasing Pressure => Decreasing Volume and Decreasing Pressure Increasing Volume.
=> Increasing Temperature => Increasing Volume and Decreasing Temperature => Decreasing Volume.
So from your problem, if pressure is increased Volume MUST 'Decrease' and if temperature is 'Increased' the volume MUST 'Increase'.
To set this up ...
V(final) = [V(initial)](Pressure Effects)(Temperature Effects)
V(initial)=V(1) --- V(final)=V(2)
P(initial)=P(1) --- P(final)=P(2)=4[P(1)]
T(initial)=T(1) --- T(final)=T(2)=3[T(1)]
V(final) = V(1)[Ratio of P(1):P(2) that will cause V(1) to 'Decrease'][Ratio of T(1):T(2) that will cause V(1) to 'Increase']
V(final) = V(1)[(P(1)/P(2)][T(2)/T(1)] = V(1)[P(1)/4P(1)][3T(1)/T(1)] = V(1)(1/4)(3/1) = (3/4)V(1) ... The final volume will be smaller than initial volume because the 'Pressure Effect' is greater than the 'Temperature Effect'.
=> Increasing Pressure => Decreasing Volume and Decreasing Pressure Increasing Volume.
=> Increasing Temperature => Increasing Volume and Decreasing Temperature => Decreasing Volume.
So from your problem, if pressure is increased Volume MUST 'Decrease' and if temperature is 'Increased' the volume MUST 'Increase'.
To set this up ...
V(final) = [V(initial)](Pressure Effects)(Temperature Effects)
V(initial)=V(1) --- V(final)=V(2)
P(initial)=P(1) --- P(final)=P(2)=4[P(1)]
T(initial)=T(1) --- T(final)=T(2)=3[T(1)]
V(final) = V(1)[Ratio of P(1):P(2) that will cause V(1) to 'Decrease'][Ratio of T(1):T(2) that will cause V(1) to 'Increase']
V(final) = V(1)[(P(1)/P(2)][T(2)/T(1)] = V(1)[P(1)/4P(1)][3T(1)/T(1)] = V(1)(1/4)(3/1) = (3/4)V(1) ... The final volume will be smaller than initial volume because the 'Pressure Effect' is greater than the 'Temperature Effect'.