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Question: The magnitude of work done (in kJ) in isothermal reversible expansion of 1 m³ of He gas ($C_v = \fra...

The magnitude of work done (in kJ) in isothermal reversible expansion of 1 m³ of He gas (Cv=32RC_v = \frac{3}{2}R) at 273.15 K and 10 atm to final pressure of 1 atm (in kJ) is -x kJ. Find the sum of numbers of x.

Answer

9

Explanation

Solution

The work done in an isothermal reversible expansion of an ideal gas is given by the formula:

W=−nRTln⁡(V2V1)W = -nRT \ln\left(\frac{V_2}{V_1}\right)

Using the ideal gas law P1V1=P2V2=nRTP_1V_1 = P_2V_2 = nRT for an isothermal process, we can write V2V1=P1P2\frac{V_2}{V_1} = \frac{P_1}{P_2}.

So, the work done can also be expressed as:

W=−nRTln⁡(P1P2)W = -nRT \ln\left(\frac{P_1}{P_2}\right)

We also know that nRT=P1V1nRT = P_1V_1. Therefore,

W=−P1V1ln⁡(P1P2)W = -P_1V_1 \ln\left(\frac{P_1}{P_2}\right)

We are given:

Initial volume V1=1 m3V_1 = 1 \text{ m}^3
Initial pressure P1=10 atmP_1 = 10 \text{ atm}
Final pressure P2=1 atmP_2 = 1 \text{ atm}
Temperature T=273.15 KT = 273.15 \text{ K}
The gas is Helium, which is an ideal gas.

To calculate the work in Joules, we need to use SI units for pressure and volume.

1 atm=101325 Pa1 \text{ atm} = 101325 \text{ Pa}.
P1=10 atm=10×101325 Pa=1013250 PaP_1 = 10 \text{ atm} = 10 \times 101325 \text{ Pa} = 1013250 \text{ Pa}.
V1=1 m3V_1 = 1 \text{ m}^3.

Now, substitute the values into the work formula:

W=−(1013250 Pa)(1 m3)ln⁡(10 atm1 atm)W = -(1013250 \text{ Pa})(1 \text{ m}^3) \ln\left(\frac{10 \text{ atm}}{1 \text{ atm}}\right)
W=−1013250 J×ln⁡(10)W = -1013250 \text{ J} \times \ln(10)

Using the value ln⁡(10)≈2.302585\ln(10) \approx 2.302585:

W≈−1013250 J×2.302585W \approx -1013250 \text{ J} \times 2.302585
W≈−2331100.8 JW \approx -2331100.8 \text{ J}

Convert the work done to kilojoules:

W≈−2331.1008 kJW \approx -2331.1008 \text{ kJ}

The question states that the magnitude of work done is -x kJ. This means the work done is W=−x kJW = -x \text{ kJ}.

So, −x kJ≈−2331.1008 kJ-x \text{ kJ} \approx -2331.1008 \text{ kJ}.
x≈2331.1008x \approx 2331.1008.

The question asks for the sum of numbers of x. Given the context of such problems, this usually refers to the sum of the digits of the integer part or the rounded value of x. The value 2331.1008 is very close to the integer 2331. Let's assume x is the rounded integer value of the magnitude.

Rounding 2331.1008 to the nearest integer gives 2331.

So, let's take x=2331x = 2331.

The sum of the numbers (digits) of x is the sum of the digits of 2331.

Sum of digits = 2+3+3+1=92 + 3 + 3 + 1 = 9.