Question

1. a pressure gauge reads 30 psig. if atmospheric pressure is 14.7 psia, what is the absolute pressure? mark only one oval. 44.7 psia 60 psia 15.3 psia 30 psia 19. in an adiabatic expansion process, what happens to the temperature of an ideal gas? mark only one oval. becomes zero remains constant increases decreases 20. during which process does a gas do the most work when expanding from the same initial to final volume? mark only one oval. isothermal isobaric isochoric adiabatic

Explanation:

Step1: Absolute pressure formula

Absolute pressure \(P_{abs}\) is the sum of gauge pressure \(P_{g}\) and atmospheric pressure \(P_{atm}\). The formula is \(P_{abs}=P_{g}+P_{atm}\).

Step2: Substitute values

Given \(P_{g} = 30\) psig and \(P_{atm}=14.7\) psia. Substitute into the formula: \(P_{abs}=30 + 14.7\).

Step3: Calculate

\(30+14.7 = 44.7\) psia.

For question 19:

In an adiabatic expansion (\(Q = 0\)), using the first - law of thermodynamics \(\Delta U=Q - W\) (where \(Q = 0\)), so \(\Delta U=-W\). For an ideal gas, \(U = nC_{v}T\). As the gas expands (\(W>0\)), \(\Delta U<0\) and since \(U\) is a function of temperature (\(U = nC_{v}T\)), temperature decreases.

For question 20:

The work done by a gas \(W=\int_{V_{1}}^{V_{2}}P dV\). For isobaric process \(P\) is constant, \(W = P\Delta V\). For isothermal \(W=nRT\ln\frac{V_{2}}{V_{1}}\), for adiabatic \(W=\frac{P_{1}V_{1}-P_{2}V_{2}}{\gamma - 1}\), and for isochoric \(W = 0\) (since \(\Delta V=0\)). Comparing the expressions (assuming same \(V_{1}\) and \(V_{2}\)), the isobaric process has the largest value of work as it is directly proportional to pressure (which is constant and non - zero for expansion) and volume change.

Answer:

1. 44.7 psia
2. Decreases
3. Isobaric