Question

1. in an isochoric process, if temperature increases, what must happen to pressure according to gas laws? mark only one oval. pressure remains constant pressure decreases pressure becomes zero pressure increases 17. a piston compresses gas in a cylinder while maintaining constant temperature. what type of work is being done on the system? mark only one oval. work increases internal energy only no work is done because temperature is constant work decreases the systems total energy work is done but internal energy remains constant 18. which statement best explains why specific volume and density are inversely related? mark only one oval. as the space occupied by each unit of mass increases, the mass per unit space decreases they measure different physical properties specific volume is mass per unit volume while density is volume per unit mass they are only related at constant temperature

Explanation:

16.

For an isochoric process, \(V=\text{constant}\). According to the ideal gas law \(PV = nRT\). When \(V\) is constant, \(P\propto T\). So if \(T\) increases, \(P\) must increase.

For an ideal gas, the internal energy \(U\) is a function of temperature only (\(U = nC_{v}T\)). Since \(T=\text{constant}\), \(\Delta U=0\). Work \(W\) is done on the gas (compression). From the first law of thermodynamics \(\Delta U=Q - W\). Since \(\Delta U = 0\), \(Q = W\). But the key is that internal energy (a function of \(T\)) remains constant while work is done.

Specific volume \(v=\frac{V}{m}\) and density \(
ho=\frac{m}{V}\), so \(v=\frac{1}{
ho}\). If \(v\) (space per unit mass) increases, \(
ho\) (mass per unit space) decreases.

Answer:

Pressure increases

17.