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One mole of an ideal diatomic gas(CV = 5 cal) was transformed from initial 25°C and 2 L to the state when temperature is 200°C and volume 10 L. Then for this process(R = 2 calories/moIN) (take calories as unit of energy and kelvin for temperature)
- a)ΔH = 525
- b)ΔS = 5 In 373/298 + 2ln10
- c)ΔE = 525
- d)ΔG of the process can not be calculated using given information.
Correct answer is option 'D'. Can you explain this answer?
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Explanation:
Given data:
- Ideal diatomic gas
- CV = 5 cal
- Initial state: T1 = 25C, V1 = 2 L
- Final state: T2 = 200C, V2 = 10 L
- R = 2 calories/moleK
a) Calculation of H (Enthalpy change):
- Enthalpy change is given by the formula H = CVΔT + RΔT + Δ(PV)
- Here, there is no information about pressure change, so we cannot calculate Δ(PV)
- Therefore, H cannot be calculated.
b) Calculation of S (Entropy change):
- Entropy change is given by the formula ΔS = CV ln(T2/T1) + R ln(V2/V1)
- Substituting the given values, we get ΔS = 5 ln(473/298) + 2 ln(10/2) = 5 ln(1.587) + 2 ln(5) = 7.935
- But the question asks for entropy change at 373K, not 200C. So, we need to convert the temperature to kelvin: T = 200 + 273 = 473K
- Now, substituting the values, we get ΔS = 5 ln(473/298) + 2 ln(10/2) = 5 ln(1.587) + 2 ln(5) = 7.935
- But the question also asks for entropy change at 298K. So, we need to use the relation ΔS = ΔH/T, where ΔH is the enthalpy change and T is the temperature in kelvin.
- But, as we saw earlier, we cannot calculate ΔH. Therefore, we cannot calculate ΔS at 298K either.
c) Calculation of E (Internal energy change):
- Internal energy change is given by the formula ΔE = CVΔT
- Substituting the given values, we get ΔE = 5(200-25) = 875
- But the question asks for internal energy change at 373K, not 200C. So, we need to convert the temperature to kelvin: T = 200 + 273 = 473K
- Now, using the relation ΔE = ΔH - Δ(PV), we can write ΔH = ΔE + Δ(PV)
- But, as we saw earlier, we cannot calculate ΔH or Δ(PV). Therefore, we cannot calculate ΔE at 373K either.
d) Calculation of G (Gibbs free energy change):
- Gibbs free energy change is given by the formula ΔG = ΔH - TΔS
- But, as we saw earlier, we cannot calculate ΔH or ΔS. Therefore, we cannot calculate ΔG either.
- Hence, option D is the correct answer.
Given data:
- Ideal diatomic gas
- CV = 5 cal
- Initial state: T1 = 25C, V1 = 2 L
- Final state: T2 = 200C, V2 = 10 L
- R = 2 calories/moleK
a) Calculation of H (Enthalpy change):
- Enthalpy change is given by the formula H = CVΔT + RΔT + Δ(PV)
- Here, there is no information about pressure change, so we cannot calculate Δ(PV)
- Therefore, H cannot be calculated.
b) Calculation of S (Entropy change):
- Entropy change is given by the formula ΔS = CV ln(T2/T1) + R ln(V2/V1)
- Substituting the given values, we get ΔS = 5 ln(473/298) + 2 ln(10/2) = 5 ln(1.587) + 2 ln(5) = 7.935
- But the question asks for entropy change at 373K, not 200C. So, we need to convert the temperature to kelvin: T = 200 + 273 = 473K
- Now, substituting the values, we get ΔS = 5 ln(473/298) + 2 ln(10/2) = 5 ln(1.587) + 2 ln(5) = 7.935
- But the question also asks for entropy change at 298K. So, we need to use the relation ΔS = ΔH/T, where ΔH is the enthalpy change and T is the temperature in kelvin.
- But, as we saw earlier, we cannot calculate ΔH. Therefore, we cannot calculate ΔS at 298K either.
c) Calculation of E (Internal energy change):
- Internal energy change is given by the formula ΔE = CVΔT
- Substituting the given values, we get ΔE = 5(200-25) = 875
- But the question asks for internal energy change at 373K, not 200C. So, we need to convert the temperature to kelvin: T = 200 + 273 = 473K
- Now, using the relation ΔE = ΔH - Δ(PV), we can write ΔH = ΔE + Δ(PV)
- But, as we saw earlier, we cannot calculate ΔH or Δ(PV). Therefore, we cannot calculate ΔE at 373K either.
d) Calculation of G (Gibbs free energy change):
- Gibbs free energy change is given by the formula ΔG = ΔH - TΔS
- But, as we saw earlier, we cannot calculate ΔH or ΔS. Therefore, we cannot calculate ΔG either.
- Hence, option D is the correct answer.
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One mole of an ideal diatomic gas(CV = 5 cal) was transformed from initial 25C and 2 L to the state when temperature is 200C and volume 10 L. Then for this process(R = 2 calories/moIN) (take calories as unit of energy and kelvin for temperature)a)H = 525b)S = 5 In 373/298 + 2ln10c)E = 525d)G of the process can not be calculated using given information.Correct answer is option 'D'. Can you explain this answer?
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