![\begin{gathered} C)V=192\text{ cubic inches} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vz53peaa9kdzks7zhx1hikh8w61vjo15v2.png)
Step-by-step explanation
Step 1
The volume V of a given mass of gas varies directly as the temperature T and inversely as the pressure P
![\begin{gathered} V=\alpha(T)/(P) \\ \text{where}\alpha\text{ is a constant} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vwtrwji2y65kxx7thtie5d5mtrw23b5qp1.png)
A measuring device is calibrated to give V = 124.8 in3 when T = 260° and P = 25 lb/in2 ,we can use this info to get the constant
![\begin{gathered} V=\alpha(T)/(P) \\ 124.8in^3=\alpha(260)/(25(lb)/(In^2)) \\ 124.8=(\alpha260)/(25) \\ (124.8\cdot25)/(260)=\alpha \\ so,\text{ } \\ \alpha=12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yuk5hcocr08eijfcqk5apz7se3o9jce0pd.png)
no, rewrite the equation
![\begin{gathered} V=\alpha(T)/(P) \\ V=12(T)/(P) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b3lqa5m0ecfsdpqj542jc1wnnoobslz3y7.png)
Step 2
What is the volume on this device when the temperature is 320° and the pressure is 20 lb/in2
replace
![\begin{gathered} V=12(T)/(P) \\ V=12(320)/(20(lb)/(In^2)) \\ V=192\text{ cubic inches} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h71mpf71kcxmeqluhf45261tneox6i36xc.png)
I hope this helps you