One mole of each gases A, B, C and D with van der Waal's constant (atm l 2 mol -2 ) 1.348, 6.823, 4.390 and 2.438, respectively, are kept separately in four different vessels of equal volumes at

Question

One mole of each gases A, B, C and D with van der Waal's constant (atm l² mol^-2) 1.348, 6.823, 4.390 and 2.438, respectively, are kept separately in four different vessels of equal volumes at identical temperature. Their pressures are observed to be P_A, P_B, P_C and P_D, respectively. On the basis of this data alone, the order of pressure of gases is (assume other van der Waal's constant to be nearly same for all gases)

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Answer:

B

Solution:

For one mole, P = (RT)/(V-b) - a/(V²); with equal V, T and nearly equal b, the larger the a, the lower the pressure. Order of a: B(6.823) > C(4.390) > D(2.438) > A(1.348). Hence P_B < P_C < P_D < P_A. Ans: (b)

TestHub · Accepted Answer

For one mole, P = (RT)/(V-b) - a/(V 2 ); with equal V, T and nearly equal b, the larger the a, the lower the pressure. Order of a: B(6.823) > C(4.390) > D(2.438) > A(1.348). Hence P B < P C < P D < P A . Ans: (b)

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