One mole of a certain ideal gas obtains an amount of heat `Q=1.60kJ` when its temperature is increased by `DeltaT=72K`, keeping its pressure constant. The vlaue of `(C_(P))/(C_(V))` for the gas is

Options

A. `1.60`
B. `1.40`
C. `1.50`
D. `1.30`

Answer

To solve the problem step by step, we will follow the outlined procedure based on the information given in the question.

### Step 1: Use the formula for heat at constant pressure
Given that the heat \( Q \) absorbed by the gas is related to the specific heat at constant pressure \( C_P \) by the equation:
\[ Q = n C_P \Delta T \]
where:
- \( Q = 1.60 \, \text{kJ} = 1600 \, \text{J} \)
- \( n = 1 \, \text{mol} \)
- \( \Delta T = 72 \, \text{K} \)

### Step 2: Rearrange the formula to find \( C_P \)
We can rearrange the equation to solve for \( C_P \):
\[ C_P = \frac{Q}{n \Delta T} \]

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ C_P = \frac{1600 \, \text{J}}{1 \, \text{mol} \times 72 \, \text{K}} = \frac{1600}{72} \]

### Step 4: Calculate \( C_P \)
Now we perform the calculation:
\[ C_P = \frac{1600}{72} \approx 22.22 \, \text{J/mol K} \]

### Step 5: Use the relation between \( C_P \) and \( C_V \)
For an ideal gas, the relationship between \( C_P \) and \( C_V \) is given by:
\[ C_P - C_V = R \]
where \( R \) is the universal gas constant, approximately \( R = 8.31 \, \text{J/mol K} \).

### Step 6: Solve for \( C_V \)
Rearranging the equation gives:
\[ C_V = C_P - R \]
Substituting the calculated value of \( C_P \):
\[ C_V = 22.22 \, \text{J/mol K} - 8.31 \, \text{J/mol K} \approx 13.91 \, \text{J/mol K} \]

### Step 7: Calculate the ratio \( \frac{C_P}{C_V} \)
Now we can find the ratio of \( C_P \) to \( C_V \):
\[ \frac{C_P}{C_V} = \frac{22.22}{13.91} \approx 1.60 \]

### Final Answer
Thus, the value of \( \frac{C_P}{C_V} \) for the gas is approximately \( 1.60 \).

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