Question

part a
what would happen to the pressure of the box if the temperature was
doubled?
a the pressure would stay the same
b the pressure would decrease
c the pressure would increase
part b
what would happen to the pressure if the volume of the box doubled?
a the pressure would decrease
b the pressure would stay the same
c the pressure would increase
part c
if the number of particles in the box were doubled, what would happen to
the pressure?
a the pressure would increase
b the pressure would stay the same
c the pressure would decrease

Explanation:

Part A

Step1: Recall the ideal gas law

The ideal gas law is \( PV = nRT \). If \( V\) (volume), \(n\) (number of moles) are constant, then \( P\propto T\) (pressure is proportional to temperature).

Step2: Analyze the effect of doubling temperature

If \(T\) is doubled (\(T_2 = 2T_1\)), from \(P=\frac{nR}{V}T\), \(P_2=\frac{nR}{V}(2T_1) = 2P_1\). So pressure increases.

Part B

Step1: Recall the ideal gas law

Using \(PV=nRT\), if \(n\), \(T\) are constant, then \(P\propto\frac{1}{V}\) (pressure is inversely proportional to volume).

Step2: Analyze the effect of doubling volume

If \(V_2 = 2V_1\), then \(P_2=\frac{nRT}{2V_1}=\frac{1}{2}P_1\). So pressure decreases.

Part C

Step1: Recall the ideal gas law

From \(PV = nRT\), if \(V\), \(T\) are constant, then \(P\propto n\) (pressure is proportional to the number of moles).

Step2: Analyze the effect of doubling the number of particles

If \(n_2=2n_1\), then \(P_2=\frac{2n_1RT}{V}=2P_1\). So pressure increases.

Answer:

Part A: C. The pressure would increase
Part B: A. The pressure would decrease
Part C: A. The pressure would increase