When the pressure exerted on 1.0L of an ideal gas is quadrupled, and the kelvin temperature is tripled,what does the volume become?
2 answers
use combined gas laws
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'.